Question

The average score for male golfers is 87 and the average score for female golfers is 102. Use these values as the population means for men and women and assume that the population standard deviation σ = 14 strokes for both. A simple random sample of 32 male golfers and another simple random sample of 36 female golfers will be taken.

a. Calculate the standard error of the mean for both male and female golfers.

b. What is the probability that the sample mean is within 3 strokes of the populations mean for the sample of male golfers?

c. What is the probability that the sample mean is within 3 strokes of the populations mean for the sample of female golfers?

d. In which case, part (b) or part (c), is the probability of obtaining a sample mean within 3 strokes of the population mean higher? Why?

Answer 1

a)

std.error = s/sqrt(n)
= 14/sqrt(32)

= 2.4749

std.error = s/sqrt(n)
= 14/sqrt(36)
= 2.3333

b)
Here, μ = 87, σ = 2.4749, x1 = 84 and x2 = 90. We need to compute P(84<= X <= 90). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (84 - 87)/2.4749 = -1.21
z2 = (90 - 87)/2.4749 = 1.21

Therefore, we get
P(84 <= X <= 90) = P((90 - 87)/2.4749) <= z <= (90 - 87)/2.4749)
= P(-1.21 <= z <= 1.21) = P(z <= 1.21) - P(z <= -1.21)
= 0.8869 - 0.1131
= 0.7738

c)
Here, μ = 102, σ = 2.3333, x1 = 99 and x2 = 105. We need to compute P(99<= X <= 105). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (99 - 102)/2.3333 = -1.29
z2 = (105 - 102)/2.3333 = 1.29

Therefore, we get
P(99 <= X <= 105) = P((105 - 102)/2.3333) <= z <= (105 - 102)/2.3333)
= P(-1.29 <= z <= 1.29) = P(z <= 1.29) - P(z <= -1.29)
= 0.9015 - 0.0985
= 0.8030

d)

For female a sample mean within 3 strokes of the population mean higher because sample size is greater