Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 73 inches and standard deviation 6 inches. in USE SALT (a) What is the probability that an 18-year-old man selected at random is between 72 and 74 inches tall? (Round your answer to four decimal places.) 0.9928 X (b) If a random sample of twenty-nine 18-year-old men is selected, what is the probability that the mean height is between 72 and 74 inches? (Round your answer to four decimal places.) 0.3232 X

Answer 1

z score normal distribution formula

z = (x - μ) / σ

(a) between 72 and 74

P(z< 72) = (72 - 73) / 6 = -0.17

P(z< 74)  = (74 - 73) / 6 = 0.17

P(-0.17 < z < 0.17) = 0.1350

-1 Area from a value (Use to compute p from Z) Value from an area (Use to compute Z for confidence intervals) 1 Specify Parameters: Mean o SD Above 1.96 Below 1.96 Between -0.17 Outside -1.96 and 0.17 and 1.96 Results: Area (probability) 0.135

(b) random sample of 21 line

z = (x - μ) / (σ/ sqrt (n))

n = 21

P(z< 72) = (72 - 73) / (6/sqrt(21)) = -0.76

P(z< 74)  = (74 - 73) / (6/sqrt(21)) = 0.76

P(-0.76 < z < 0.76) = 0.5527

Area from a value (Use to compute p from Z) Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Mean o ܝ SD Above 1.96 Below 1.96 Between -0.76 Outside -1.96 and 0.76 and 1.96 Results: Area (probability) 0.5527