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

you are saying it as if a fact, not a question, but yes, you would be correct? ; )

step-by-step explanation:

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)