Question

16) The mean time spent solving this problem is 4.25 minutes with a standard deviation of 1.7 minutes. Find the probability a sample of 32 test takers spends a mean more than 5 minutes on this problem.

17. On average, people spend 34 minutes complaining on Mondays. The standard deviation is 3.67 minutes. Find the probability a random sample of 48 people spend a mean less than 35 minutes complaining on Mondays.

18. 27% of all women’s hair has split ends. In a sample of 145 women’s hairs find the probability more than 29% are split. Can we assume normal here?

19. Find a 98% confidence interval for the true mean length of eyelashes, if a sample 33 eyelashes has a mean length of 13.3 millimeters, with a standard deviation of 3.6.

Answer 1

16)

Here, μ = 4.25, σ = 0.3005 and x = 5. We need to compute P(X >= 5). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (5 - 4.25)/0.3005 = 2.5

Therefore,
P(X >= 5) = P(z <= (5 - 4.25)/0.3005)
= P(z >= 2.5)
= 1 - 0.9938 = 0.0062

17)
Here, μ = 34, σ = 0.5297 and x = 35. We need to compute P(X <= 35). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (35 - 34)/0.5297 = 1.89

Therefore,
P(X <= 35) = P(z <= (35 - 34)/0.5297)
= P(z <= 1.89)
= 0.9706

18)

Here, μ = 0.27, σ = 0.0369 and x = 0.29. We need to compute P(X >= 0.29). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.29 - 0.27)/0.0369 = 0.54

Therefore,
P(X >= 0.29) = P(z <= (0.29 - 0.27)/0.0369)
= P(z >= 0.54)
= 1 - 0.7054 = 0.2946

19)

sample mean, xbar = 13.3
sample standard deviation, s = 3.6
sample size, n = 33
degrees of freedom, df = n - 1 = 32

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.449

ME = tc * s/sqrt(n)
ME = 2.449 * 3.6/sqrt(33)
ME = 1.535

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (13.3 - 2.449 * 3.6/sqrt(33) , 13.3 + 2.449 * 3.6/sqrt(33))
CI = (11.8 , 14.8)