THE NORMALDISTRIBUTION

To calculate probabilities associated with normal distributionswe use the standard normal distribution.

If X is a normal distribution with mean and standarddeviation , we write X N ( , 2).

To standardise this we use the transformation: Z = X –

The standardised score tells us the number of standard deviations that a value is above the mean.

Statistical tables are used to evaluate the probabilities.

Since only positive values of z are listed in the tables, we use thefact that the normal distribution curve is symmetrical.

A clear diagram is essential for these problems.

Then, Z N (0 , 1 ).

Example 1: The random variable X has a normal distribution with mean 20 and variance 16. Find the probability that X > 26.

We have: X ~ N ( 20, 16 )

z0 z1

z1 =

[ From the tables, z = 1.5, p = 0.9332 ]

= 1.5

2620 x

26 – 20 16

= 0.0668

Now find the standardised score:

This means that 26 is 1.5 standard deviations above the mean.

P( X > 26 ) = 1 – 0.9332

Z = X –

Using:

Example 2: The random variable X has a normal distribution with mean 40 and standard deviation 12. Find P ( X > 25 ).

z0z1

4025 x

We have: X ~ N ( 40, 122 )

Now find the standardised score:

z1 = 25 – 40 12

= – 1.25

[ From the tables, z = 1.25, p = 0.8944 ]

By symmetry, the required probability is the same:

i.e. P ( X > 25 ) = 0.8944

Z = X –

Using:

Example 3: The weight, X grams, of soup put in a can by a machine is normally distributed with a mean of 180 g and a variance of 38 g2. A can is selected at random. Find the probability that the can contains less than 171 g.

We have: X ~ N ( 180, 38 )171 180 x

z 0 z1Now find the standardised score:

171 –180 38

z1 = = – 1.46

[ From the tables, z = 1.46, p = 0.9279 ]

By symmetry, the required probability is 1 – 0.9279

i.e. P ( X < 171 ) = 0.0721

Z = X –

Using:

Example 4: The distances travelled to work, X km, by the employees at a factory are normally distributed with X N ( 20, 32 ).

Find the probability that a randomly selected employee has a journey to work of between 17 km and 25 km.

zz2 z1020 x 17 25

25 – 20 32

z1 =

17 – 20 32

z2 =

= 0.88

= – 0.53

[ From the tables, z = 0.88, p = 0.8106 ]

[ From the tables, z = 0.53, p = 0.7019 ]

A1A2

A1 =

A2 =

0.3106

0.2019

The probability of between 17 and 25 km = 0.3106 + 0.2019 = 0.5125

We have: X N ( 20, 32 ).

Example 5: A machine fills jars with coffee. The weight W g of the coffee in the jar can be modelled by a normal distribution with mean 200 g and standard deviation 15 g. Find the weight of coffeethat is exceeded by 3% of the jars.

We have: W N ( 200, 152 ).

[ From the tables, p = 0.97, z = 1.88 ]

x200 x1 z0 1.88

i.e. 3% of jars contain more than 228.2 g of coffee.

Top 3%

97%

x1 = 228.2

x1 – 200 15

1.88 =

Example 6: The heights of a population of women are normally distributed with mean cm and standard deviation cm. It is known that 4% of the women are taller than 170 cm and 2.5% are shorter than 150 cm. Find the values of and .

We have: X N ( , 2).

170 x150 z 0 z1 z2

[ From the tables, p = 0.96, z = 1.75 ]

[ From the tables, p = 0.975, z = 1.96 ]

There are 96% of the women below 170cm.

There are 97.5% of the women above 150cm.

170 –

1.75 =

– 1.96 = 150 –

We now have two simultaneous equations: 1.75 = 170 – . . . (1)– 1.96 = 150 – . . . (2)

This gives = 5.39 and = 160.6

Summary of key points:

This PowerPoint produced by R.Collins ; Updated Mar. 2012

If X is a normal distribution with mean and standarddeviation , we write X N ( , 2).

To standardise this we use the transformation: Z = X –

A clear diagram is essential for these problems.

The standardised score tells us the number of standard deviations that a value is above the mean.

Note, the variance is written here