One father is 70 inches tall and the residual for his son’s height is 2.5. What is the son’s actual height? 71.7 inches
74.2 inches
76.7 inches
82.3 inches

C

Step-by-step explanation:

Since the heights of the men are assumed to be normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = heights of the men.

µ = mean height

σ = standard deviation

From the information given,

µ = 70.7 inches

σ = 2.1 inches

The probability that they have a mean height greater than 71.7 inches is expressed as

P(x > 71.7) = 1 - P(x ≤ 71.7)

For x = 71.7

Since n = 36, then

z = (71.7 - 70.7)/2.1/√36 = 2.86

Looking at the normal distribution table, the probability corresponding to the z score is 0.998

P(x > 71.7) = 1 - 0.998 = 0.002

the olimpics is not going to wokk out

step-by-step explanation: