Question

Annual votes for a small political party are approximately normally distributed with a mean of 50 000 and a standard deviation of 20 000.

a. What number of total votes would be needed by a party to fall in the lowest 30%?

b. What is the probability of having an above average vote range of between R60000 to R80000?

Answer 1

Solution:

Given: Annual votes for a small political party are approximately normally distributed with a mean of 50 000 and a standard deviation of 20 000.

Mean =

Standard Deviation =

Part a) Find x = number of total votes would be needed by a party to fall in the lowest 30%

That is find x such that:

P( X < x ) = 30%

That is:

P( X < x ) = 0.30

Thus first find z such that:

P( Z < z ) = 0.3000

Look in z table for Area = 0.3000 or its closest area and find corresponding z value.

Area = 0.3015 closest to area = 0.3000 and it corresponds to -0.5 and 0.02

thus z = -0.52

Now use following formula to find x value:

Thus the number of total votes would be needed by a party to fall in the lowest 30% are = 39600 votes.

Part b) Find the probability of having an above average vote range of between R60000 to R80000

That is find:

P( 60000 < X < 80000) = ...............?

Find z scores for x = 60000 and for x = 80000

z score formula is:

Thus we get:

P( 60000 < X < 80000) = P( 0.50 < Z < 1.50 )

P( 60000 < X < 80000) = P( Z < 1.50 ) - P( Z < 0.50 )

Look in z table for z = 1.5 and 0.00 as well as for z = 0.5 and 0.00 and find corresponding area.

P( Z < 0.50) = 0.6915

P( Z< 1.50) = 0.9332

Thus

P( 60000 < X < 80000) = P( Z < 1.50 ) - P( Z < 0.50 )

P( 60000 < X < 80000) = 0.9332 - 0.6915

P( 60000 < X < 80000) = 0.2417