Assume the heights in a male population are normally distributed with mean 70.3 inches and standard deviation 4.1 inches. If a randomly selected male’s height is at the 75th percentile, how tall is he (to the nearest tenth of an inch)?

73.1 inches

Step-by-step explanation:

We solve for this using z score formula.

z-score is z = (x-μ)/σ,

where x is the raw score,

μ is the population mean

σ is the population standard deviation.

x = unknown = randomly selected male's height

μ = 70.3 inches

σ = 4.1 inches

z = is not given but we are given its percentile which is 75

The z score for a 75th percentile = 0.674

Hence,

0.674 = x - 70.3/4.1

Cross Multiply

0.674 × 4.1 = x - 70.3

2.7634 = x - 70.3

x = 70.3 + 2.7634

x = 73.0634 inches

The height of the randomly selected male = 73.0634

Therefore, the height of the randomly selected male to the nearest tenth of an inch is 73.1 inches

it would take 7/10 of an hour too fill the pool.

step-by-step explanation:

28

step-by-step explanation:

because that's the answer