Assume that the average driving speed on the highway forms a normal distribution

Question

Assume that the average driving speed on the highway forms a normal distribution with mean = 60 mph and standard deviation =20. If you randomly select a sample of 16 drivers

a) What is the probability that the average driving speed for the sample will be greater than 50 mph?

b) What is the proportion of samples of this size (n=16) that would produce an average driving speed of 56 mph or slower?

c) What is the probability that the average driving speed of the sample will be within 5 points of the population mean?

Explanation / Answer

mean = 60 mph

standard deviation =20

sample size n= 16

standard deviation of average of sample = standard deviation/n = 20/16 = 20/4 = 5

a)

z value for average = 50 = (50-60)/20 = -0.5, correspoding p value using z-table is 0.3085

P(average < 50 ) = 0.3085

P(average > 50 ) = 1-0.3085 = 0.6915

probability that the average driving speed for the sample will be greater than 50 mph is 0.6915

b)

z value for average = 56 = (56-60)/20 = -0.2, correspoding p value using z-table is 0.4207

P(average < 56 ) = 0.4207

proportion of samples of this size (n=16) that would produce an average driving speed of 56 mph or slower is 0.4207

c)

probability that the average driving speed of the sample will be within 5 points of the population mean = P(mean-5<X<mean+5) i.e P(55<X<65)

z value for 55 is (55-60)/5 = -1 , correspoding p value using z table is 0.1587

P(X<55) = 0.1587

z value for 65 is (65-60)/5 = 1 , correspoding p value using z table is 0.8413

P(X<65) = 0.8413

P(55<X<65) = 0.8413-0.1587 = 0.6826

probability that the average driving speed of the sample will be within 5 points of the population mean is 0.6826

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