Edexcel International AS Maths: Statistics 1

Topic Questions

3.2 Normal Distribution

1a
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1 mark

A continuous random variable can take any value within a given range.  Many naturally occurring continuous quantities can be modelled using the Normal Distribution, for example the height of human beings; the mass of new born puppies or the distribution of all A Level maths exam results.

Give a different example of a quantity that could be modelled using the normal distribution.

1b
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3 marks

The graph of the normal distribution has a characteristic bell shape that is symmetrical about the mean, mu.  If X has a normal distribution with a mean,mu , and variance,sigma squared,  then it can be written as X tilde space N left parenthesis mu comma sigma squared right parenthesis.

 For X tilde N left parenthesis mu comma sigma squared right parenthesis, state:

(i)
straight P left parenthesis X space less than space mu right parenthesis

(ii)
straight P left parenthesis X space greater or equal than space mu right parenthesis

(iii)
straight P left parenthesis X space equals space mu right parenthesis
1c
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1 mark

Using your answers to part (b), or otherwise, explain why there is no difference between greater or equal than and greater than or less or equal than and less than when calculating normal probabilities.

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2
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5 marks

For the standard normal distribution, Z tilde straight N left parenthesis 0 comma 1 ² right parenthesis , the probability  straight P left parenthesis Z less or equal than space z right parenthesis can be written as capital phi left parenthesis z right parenthesis.

The value of capital phi left parenthesis 0.23 right parenthesis space would be found in the table by looking up z equals 0.23 to get the answer capital phi left parenthesis 0.23 right parenthesis equals 0.5910.

Use the table of z-values in the formula booklet to write down the value of:

(i)
capital phi left parenthesis 0.04 right parenthesis
(ii)
capital phi left parenthesis 0.63 right parenthesis
(iii)
capital phi left parenthesis 1 right parenthesis
(iv)
capital phi left parenthesis 1.59 right parenthesis
(v)
capital phi left parenthesis 3.30 right parenthesis

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3a
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4 marks

For the standard normal distribution,  Z tilde N left parenthesis 0 comma 1 squared right parenthesis, the probability P left parenthesis Z less or equal than space z right parenthesis can be written as capital phi left parenthesis z right parenthesis.

To solve space capital phi left parenthesis z right parenthesis equals 0.6808  we would find the value 0.6808 in a column with heading capital phi left parenthesis z right parenthesis and read off the corresponding z-value, so  capital phi left parenthesis 0.47 right parenthesis equals 0.6808.

Use the table of z-values in the formula booklet to solve:

(i)
capital phi left parenthesis z right parenthesis equals 0.5120
(ii)
capital phi left parenthesis z right parenthesis equals 0.9147
(iii)
capital phi left parenthesis z right parenthesis equals 0.8643
(iv)
capital phi left parenthesis z right parenthesis equals 0.9783.
3b
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2 marks

The formula booklet also contains a table of percentage points of z  such that P left parenthesis Z greater than z right parenthesis equals p.  From this table we can see that if space P left parenthesis Z greater than z right parenthesis equals 0.0005 spacethen space z equals 3.2905.

Use the table of percentage points to find the value of z such that:

(i)
P left parenthesis Z greater than z right parenthesis equals 0.2
(ii)
P left parenthesis Z greater than z right parenthesis equals 0.001.

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4a
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4 marks

For the standard normal distribution, Z tilde N left parenthesis 0 comma 1 ² right parenthesis, the probability P left parenthesis Z less or equal than space z right parenthesis can be written as capital phi left parenthesis z right parenthesis.

To calculate P left parenthesis Z greater than z right parenthesis the following formulae can be used:

straight P left parenthesis Z greater than z right parenthesis equals 1 minus straight P left parenthesis Z less than z right parenthesis comma space space space straight P left parenthesis Z greater than z right parenthesis equals 1 minus capital phi left parenthesis z right parenthesis and straight P left parenthesis Z greater than z right parenthesis equals capital phi left parenthesis negative z right parenthesis.

Use the formulae above to write each of the following in the form 1 minus capital phi left parenthesis z right parenthesis or space capital phi left parenthesis negative z right parenthesis and then use the table of z-values to find its value:

(i)
P left parenthesis Z greater than 0.74 right parenthesis
(ii)
P left parenthesis Z greater than negative 0.74 right parenthesis
(iii)
P left parenthesis Z greater than 2.14 right parenthesis
(iv)
P left parenthesis Z greater than negative 1.52 right parenthesis
4b
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5 marks

Use the formulae above to rewrite each equation in the form space capital phi left parenthesis z right parenthesis equals pcapital phi left parenthesis negative z right parenthesis equals p or P left parenthesis Z greater than z right parenthesis and then use the table of z-values and the table of percentage points to find the value of z such that:

(i)
P left parenthesis Z greater than z right parenthesis equals 0.9306
(ii)
P left parenthesis Z greater than z right parenthesis equals 0.0694
(iii)
P left parenthesis Z greater than z right parenthesis equals 0.1292
(iv)
P left parenthesis Z less than z right parenthesis equals 0.995.

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5a
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3 marks

For the standard normal distribution,  Z tilde N left parenthesis 0 comma 1 ² right parenthesis, the probability P left parenthesis Z less or equal than space z right parenthesis can be written as capital phi left parenthesis z right parenthesis.

The table of z-values does not contain negative z-values. To calculate probabilities involving negative z-values the following formulae can be used:

 capital phi left parenthesis negative z right parenthesis equals 1 minus capital phi left parenthesis z right parenthesis  and   P left parenthesis Z greater than negative z right parenthesis equals capital phi left parenthesis z right parenthesis.

Use the table of z -values in the formula booklet and the formula above to find the value of:

(i)
capital phi left parenthesis negative 0.5 right parenthesis
(ii)
capital phi left parenthesis negative 0.58 right parenthesis
(iii)
capital phi left parenthesis negative 2.75 right parenthesis
5b
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5 marks

Use the formula above to find the value of capital phi left parenthesis z right parenthesis  and then use the table of z-values and the table of percentage points in the formula booklet to find the value of negative z  and hence solve the following.  Where possible give the value for z with the greater number of decimal places.

(i)
capital phi left parenthesis z right parenthesis equals 0.1093
(ii)
capital phi left parenthesis z right parenthesis equals 0.1500
(iii)
capital phi left parenthesis z right parenthesis equals 0.0548
(iv)
capital phi left parenthesis z right parenthesis equals 0.0005

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6a
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4 marks

For the standard normal distribution, Z tilde N left parenthesis 0 comma 1 ² right parenthesis, the probability P left parenthesis Z less or equal than space z right parenthesis can be written as capital phi left parenthesis z right parenthesis.

To calculate the probability between two values the following formula can be used:

P left parenthesis a less than Z less than b right parenthesis equals capital phi left parenthesis b right parenthesis minus capital phi left parenthesis a right parenthesis.

Use the formula above to rewrite the following in the form space capital phi left parenthesis b right parenthesis minus capital phi left parenthesis a right parenthesis and then use the table of z-values to calculate the probability and round your answer to three decimal places:

(i)
P left parenthesis 0.12 less than Z less than 1.34 right parenthesis
(ii)
P left parenthesis negative 2 less than Z less than 1 right parenthesis
(iii)
P left parenthesis negative 1.5 less than Z less than negative 0.5 right parenthesis
6b
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4 marks

Use the formula above to rewrite the following in the form space capital phi left parenthesis b right parenthesis minus capital phi left parenthesis a right parenthesis equals p  and then use the table of z-values and the table of percentage points to find the value of z such that:

(i)
P left parenthesis 0 less than Z less than z right parenthesis equals 0.4505
(ii)
P left parenthesis z less than Z less than 1.3 right parenthesis equals 0.9027
(iii)
P left parenthesis negative z less than Z less than z right parenthesis equals 0.6826

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7a
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4 marks

For the standard normal distribution, Z tilde N left parenthesis 0 comma 1 ² right parenthesis, the probability P left parenthesis Z space less than space z right parenthesis can be written as capital phi left parenthesis z right parenthesis.

A random variable X tilde N left parenthesis mu comma sigma squared right parenthesis can be coded to Z tilde N left parenthesis 0 comma 1 squared right parenthesis comma using the formula:

space Z equals fraction numerator left parenthesis X minus mu right parenthesis over denominator sigma end fraction

For the random variable X tilde space N left parenthesis 21 comma 4 squared right parenthesis,

(i)
write down the values of mu and sigma,
(ii)
calculate the z-value that corresponds to space X equals 26,
(iii)
write space P left parenthesis X greater than 26 right parenthesis in the form capital phi left parenthesis z right parenthesis,
(iv)
calculate space P left parenthesis X greater than 26 right parenthesis.
7b
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6 marks

For the random variable X space tilde space N left parenthesis 54 comma 5 squared right parenthesis, write in terms of capital phi left parenthesis z right parenthesis and hence find:

(i)
P left parenthesis X less or equal than 60 right parenthesis
(ii)
P left parenthesis X less than 51 right parenthesis
(iii)
P left parenthesis X greater or equal than 58 right parenthesis.
7c
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4 marks

For space X space tilde space N left parenthesis 187 comma 100 right parenthesis,

(i)
explain why space sigma equals 10,
(ii)
write space P left parenthesis 190 less than X less than 203 right parenthesis in the form space P left parenthesis a less than Z less than b right parenthesis,
(iii)
hence calculatespace P left parenthesis 190 less than X less than 203 right parenthesis.

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1a
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5 marks

The test scores, X, of a group of RAF recruits in an aptitude test are modelled as a normal distribution with X tilde N left parenthesis 210 comma 27.8 squared right parenthesis.

(i)
Find the values of a and b such that P left parenthesis X less than a right parenthesis equals 0.25 and P left parenthesis X greater than b right parenthesis equals 0.25.
(ii)
Hence find the interquartile range of the scores.
1b
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3 marks

Those who score in the top 30% on the test move on to the next stage of training.

One of the recruits, Amelia, achieves a score of 231. Determine whether Amelia will move on to the next stage of training.

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2a
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8 marks

The random variable X tilde N left parenthesis 13 comma space 15 right parenthesis

Find the value of a, to 3 significant figures, such that:

(i)
P left parenthesis X greater than a right parenthesis equals 0.2
(ii)
P left parenthesis a less or equal than X less or equal than 14 right parenthesis equals 0.5
(iii)
P left parenthesis 13 minus a less or equal than X less or equal than 13 plus a right parenthesis equals 0.9372
2b
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2 marks

Explain why there are no values of a such that space P left parenthesis 14 less or equal than X less or equal than a right parenthesis equals 0.5.

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3a
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3 marks

The standard normal distribution is Z tilde space N left parenthesis 0 comma space 1 squared right parenthesis.

(i)
Write down, to 4 decimal places, the value of z for which P left parenthesis Z greater than z right parenthesis equals 0.1.
(ii)
Use your answer to part (a)(i) along with the properties of the normal distribution to work out the values of a and b for which P left parenthesis Z less than a right parenthesis equals 0.9 and P left parenthesis Z less than b right parenthesis equals 0.1.

3b
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2 marks

The weights, W kg, of coconuts grown on the Coconutty As They Come coconut plantation are modelled as a normal distribution with mean 1.25 kg and standard deviation 0.38 kg.  The plantation only considers coconuts to be exportable if their weights are greater than 10% but less than 90%.

Use your answer to part (a)(ii) to find the range of possible weights, to the nearest 0.01 kg, for an exportable coconut.

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4a
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4 marks

The random variable X tilde N left parenthesis mu comma sigma squared right parenthesis.  It is known that P left parenthesis X greater than 34.451 right parenthesis equals 0.001 and P left parenthesis X less than 14.792 right parenthesis equals 0.1977.

(i)
Use the relationship between X and the standard normal variable Z to show that the following equation must be true:

mu plus 3.0902 sigma equals 34.451

(ii)
Write down a second equation in terms of mu  and sigma .
4b
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2 marks

By solving the simultaneous equations in (a), determine the values of mu and sigma

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5a
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3 marks

A machine is used to fill bottles of a particular brand of soft drink.  The volume, V ml, of soft drink in the bottles is normally distributed with mean 450 ml and standard deviation  ml.  Given that 5% of the cans contain less than 429 ml of soft drink, find:

the value of sigma,

5b
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3 marks

P left parenthesis V greater or equal than 475 right parenthesis correct to 3 decimal places,

5c
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2 marks

the probability that the volume of a randomly selected bottle is no more than one standard deviation away from the mean.

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6a
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3 marks

Paul enjoys solving sudoku puzzles. The lengths of time he spends on sudokus in a week are normally distributed with a mean of 2048 minutes and a standard deviation of 64 minutes.

Find the probability that in a given month Paul spends less than 1945 minutes solving sudoku puzzles.

6b
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4 marks

Estimate the number of weeks in a year that Paul spends between 2019 and 2091 minutes solving sudoku puzzles.

6c
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4 marks

Assuming it takes Paul exactly 10 minutes to solve any sudoku puzzle, find the greatest number, n, of sudoku puzzle such that the probability of Paul solving less than n puzzles in a week is less than 0.01.

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7a
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5 marks

A machine is used to fill bags of potatoes for a supermarket chain. The weight, W kg, of potatoes in the bags is normally distributed with mean 3 kg and standard deviation sigma kg. 

Given that 7% of the bags contain a weight of potatoes that is at least 50 g more than the mean, find:

P left parenthesis 2.9 less or equal than W less or equal than 3.1 right parenthesis.

7b
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4 marks

Two of the bags of potatoes are chosen at random.

Find the probability that neither of the bags will contain less than 2.96 kg of potatoes.

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1a
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2 marks

For the random variable,space space X space tilde space N left parenthesis 32 comma 9 right parenthesis,

(i)
Write down the mean and standard deviation.
(ii)
Draw a sketch of the graph and clearly label the mean. 
1b
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2 marks
(i)
With the help of your diagram, explain how you should know, without carrying out any calculations, that Pleft parenthesis X less or equal than 34 right parenthesis greater than 0.5. 
(ii)
With the help of your diagram, explain how you should know, without carrying out any calculations, that Pleft parenthesis X greater or equal than 38 right parenthesis less than 0.5.
1c
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6 marks

Use the formula Z equals fraction numerator X minus mu over denominator sigma end fraction to find appropriate z-values and then use the table of z-values to calculate, to three decimal places,

(i)
P left parenthesis X less or equal than space 34 right parenthesis
(ii)
P left parenthesis X greater or equal than 38 right parenthesis
(iii)
P left parenthesis 34 space less or equal than X space less or equal than space 38 right parenthesis.

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2a
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4 marks

For the standard normal distribution Z tilde N left parenthesis 0 comma 1 squared right parenthesis, use the table of z-values to find:

(i)
P left parenthesis Z less than 1.5 right parenthesis
(ii)
P left parenthesis Z greater than negative 0.8 right parenthesis
(iii)
P left parenthesis negative 2.1 less than Z less than negative 0.3 right parenthesis
2b
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3 marks

The random variable X tilde N left parenthesis 2 comma 0.1 squared right parenthesis.

Using the coding relationship between X and Y, find the values of a comma space b comma space c and d such that:

(i)
P left parenthesis X less than a right parenthesis equals P left parenthesis Z less than 1.5 right parenthesis
(ii)
P left parenthesis X greater than b right parenthesis equals P left parenthesis Z greater than negative 0.8 right parenthesis
(iii)
P left parenthesis c less than X less than d right parenthesis equals P left parenthesis negative 2.1 less than Z less than negative 0.3 right parenthesis

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3a
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3 marks

The random variable X space tilde space N left parenthesis 50 comma 49 right parenthesis

(i)
Write down the values of mu and sigma .

(ii)
Use the percentage points table to find the value of z such that space P left parenthesis Z greater or equal than z right parenthesis equals 0.025.

(iii)
Use the formula Z equals fraction numerator X minus mu over denominator sigma end fraction to find the value of a such that space P left parenthesis X greater or equal than a right parenthesis equals 0.025.
3b
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2 marks
(i)
Find the value of z, to 4 decimal places, such that P left parenthesis Z less or equal than z right parenthesis equals 0.999.

(ii)
Hence, use the formula Z equals fraction numerator X minus mu over denominator sigma end fraction to find the value of b such that space P left parenthesis X less or equal than b right parenthesis equals 0.999.

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4a
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4 marks

For the random variable space X space tilde space N left parenthesis mu comma 36 right parenthesis it is known that space P left parenthesis X less than 51 right parenthesis equals 0.8665.

(i)
Write down the value of sigma.

 

(ii)
Calculate the value of z such that space P left parenthesis Z less than z right parenthesis equals 0.8665.

 

(iii)
Use the formula Z equals fraction numerator X minus mu over denominator sigma end fraction to show that space 51 minus mu equals 6.66.

 

(iv)
Hence calculate the value of mu.
4b
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3 marks

For the random variable Y tilde N left parenthesis 93 comma space sigma squared right parenthesis it is known that space P left parenthesis Y greater than 101 right parenthesis equals 0.0735.

(i)
Calculate the value of z such that space P left parenthesis Z greater than z right parenthesis equals 0.0735.

 

(ii)
Use the formula Z equals fraction numerator X minus mu over denominator sigma end fraction to show that space 1.45 sigma equals 8.

 

(iii)
Hence calculate the value of sigma.

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5a
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5 marks

For the random variable X tilde N left parenthesis 23 comma space 4 squared right parenthesis find the following probabilities:

(i)
P left parenthesis X less than 20 right parenthesis
(ii)
P left parenthesis X greater or equal than 29 right parenthesis
(iii)
P left parenthesis 20 less or equal than X less than 29 right parenthesis
5b
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5 marks

For the random variable Y tilde N left parenthesis 100 comma space 225 right parenthesis find the following probabilities:

(i)
P left parenthesis Y less or equal than 90 right parenthesis
(ii)
P left parenthesis Y greater than 145 right parenthesis
(iii)
P left parenthesis 85 less or equal than Y less or equal than 115 right parenthesis

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6a
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5 marks

The random variable X tilde N left parenthesis 10 comma space 9 right parenthesis.

Find the value of a and the value of b, each to 2 decimal places, such that:

(i)
P left parenthesis X less than a right parenthesis equals 0.4
(ii)
P left parenthesis X greater than b right parenthesis equals 0.15
6b
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2 marks

Use a sketch of the distribution of X to explain why P left parenthesis a less or equal than X less or equal than b right parenthesis equals 0.45.

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7a
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2 marks

The random variable X tilde N left parenthesis mu comma space sigma squared right parenthesis.
It is known that P left parenthesis X greater than 36.88 right parenthesis equals 0.025 and P left parenthesis X less than 27.16 right parenthesis equals 0.1

Find the values of a and b for which P left parenthesis Z greater than a right parenthesis equals 0.025 and  P left parenthesis Z less than b right parenthesis equals 0.1,  where Z is the standard normal variable. Give your answers correct to 4 decimal places.

7b
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2 marks

Use your answers to part (a), along with the relationship between Z and X, to show that the following simultaneous equations must be true:

mu plus 1.96 sigma equals 36.88

mu minus 1.2816 sigma equals 27.16

7c
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2 marks

By solving the simultaneous equations in (b), determine the values of mu and sigma. Give your answers correct to 2 decimal places.

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1a
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4 marks

The test scores, X, of a group of Royal Navy recruits in an aptitude test are modelled as a normal distribution with  X tilde N left parenthesis 520 comma space 89.9 squared right parenthesis.

Find the interquartile range of the scores.

1b
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5 marks

Those who get a score over 670 are offered a role within the submarine service.  Only those who get a score over 750 are offered a leadership role within the submarine service.

Given that Mervyn, one of the recruits, has been offered a role within the submarine service, find the probability that he is offered a leadership role.

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2a
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4 marks

The distribution of the test scores, X, of a group of British Army officer cadets on an aviation aptitude test is modelled as a normal distribution with  X tilde N left parenthesis 120 comma space 26.5 squared right parenthesis.

Only cadets who score in the top 10% on the test are eligible to proceed directly to helicopter pilot training.  Cadets whose scores are between the 40th and 90th percentiles, however, are eligible to resit the test in an attempt to improve their scores.

Given that it is only possible to receive an integer number of marks as a score on the test, determine the range of test scores for which cadets would be eligible to resit the test.

2b
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4 marks
(i)
Find P left parenthesis X greater than 180 right parenthesis.
(ii)
The maximum score it is possible to receive on the test is 180.  Use this fact, and your answer to part (b)(i), to criticise the model being used for the score distribution.

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3
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6 marks

The weight, W kg, of the feed in a sack of pheasant feed produced by a certain manufacturer is modelled as W tilde N open parentheses 20 comma space 1 over 3600 close parentheses.

Roger buys three sacks of the manufacturer’s pheasant feed to feed to the pheasants who have begun showing up at his backyard bird feeding station.

Find the probability that all three sacks contain feed with a weight that is within 35 g of 20 kg.

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4a
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6 marks

The masses of turkeys can be modelled using a normal distribution with a mean of 4.7 kg and a variance of 1.9 kg².  Nicholas, a farmer, classes a turkey as ‘holiday ready’ if it weighs more than 5.5 kg.

A turkey is selected at random. Given that it is ‘holiday ready’, find the probability that it weighs less than 6 kg.

4b
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5 marks

Nicholas takes a large sample of ‘holiday ready’ turkeys, estimate the median mass of the turkeys in his sample.

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5a
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6 marks

An archaeologist has devoted his life to studying ancient Greek vases produced by a particular Boeotian pottery workshop.  The vases were made to a standard pattern, and after measuring a very large number of them the archaeologist has found that 5% of the vases have a mass greater than 2.237 kg, while only 1% of them have a mass less than 1.906 kg.

Given that the masses of the vases may be assumed to be distributed normally, find the mean and standard deviation of the distribution.

5b
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4 marks

The archaeologist has found that vases made by the workshop with a mass less than m kg are particularly fragile and require special care.  A museum has just purchased a collection of  vases produced by the workshop.  The probability of all 10 vases requiring special care is 9.766 cross times 10 to the power of 24.

Find the value of m.

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6a
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5 marks

The Monkey Puzzle Tree Marketing Board has proposed a new scheme that will measure the puzzlingness of monkey puzzle trees in terms of a unit called the ‘fuddle’. It is found that the puzzlingness measurements, X, of monkey puzzle trees grown on the We ♥ Puzzling Monkeys monkey puzzle tree plantation can be modelled as a normal distribution with mean mu fuddles and standard deviation sigma fuddles.  Because the owners of the plantation are committed to making sure that the monkey puzzle trees they sell to gardeners are neither too puzzling nor not puzzling enough, the plantation only considers monkey puzzle trees to be saleable if their puzzlingness is between the 10% and 97.5% percentiles of the plantation’s monkey puzzle tree puzzlingness measurements.

Find the range of possible puzzlingness measurements for a saleable monkey puzzle tree. Your answer should be given in terms of mu and sigma.

6b
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3 marks

Given that space P left parenthesis X less than mu minus 8 right parenthesis equals 0.2  find space P left parenthesis X greater than mu plus 8 │ X greater than mu minus 8 right parenthesis.

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