Question

Question 1)

Assume that the heights of American men are normally distributed with a mean of 69.2 inches and a standard deviation of 3.2 inches.

What is the probability that a randomly selected man will be between 5'9" and 6'1" tall? (Round your answer to four decimal places.)

Question 2)

Answer the question for a normal random variable x with mean μ and standard deviation σ specified below. (Round your answer to one decimal place.)

μ = 36 and σ = 11.

Find a value of x that has area 0.01 to its right.

Question 3)

ASK YOUR TEACHER

Let z be a standard normal random variable with mean μ = 0 and standard deviation σ = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.)

to the right of −1.85

Question 4)

Let z be a standard normal random variable with mean μ = 0 and standard deviation σ = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.)

to the left of 1.1

Answer 1

Solution:-

1) Given that mean = 69.2, sd = 3.2

=> 5.9" = 70.8 inches, 6.1" = 73.2, Therefore 1 feet = 12inches

P(70.8 < X < 73.2) = P((70.8-69.2)/3.2 < (X-mean)/sd < (73.2-69.2)/3.2)
= P(0.5 < Z < 1.25)
= 0.2029

2) Given that μ = 36 and σ = 11.

P(X > x0) = 0.01, Z = 2.33

X = μ + Z*σ = 36 + 2.33*11 = 61.63

3) => mean μ = 0 and standard deviation σ = 1

P(Z > −1.85) = P(Z < 1.85) = 0.9678

4) P(Z < 1.1) = 0.8643