Question

1.The manufacturer of cans of salmon that are supposed to have a net weight of 6 ounces tells you that the net weight is actually a normal random variable with a mean of 5.98 ounces and a standard deviation of 0.12 ounce. Suppose that you draw a random sample of 44 cans. Find the probability that the mean weight of the sample is less than 5.94 ounces.

Probability =

2.Scores for men on the verbal portion of the SAT-I test are normally distributed with a mean of 509and a standard deviation of 112.
(a)  If 1 man is randomly selected, find the probability that his score is at least 588.

(b)  If 18 men are randomly selected, find the probability that their mean score is at least 588.

Answer 1

Question 1

Given

µ = 5.98

σ = 0.12

n = 44

We have to find P(Xbar<5.94)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (5.94 – 5.98)/[0.12/sqrt(44)]

Z = -0.04/ 0.018091

Z = -2.21104

P(Z<-2.21104) = P(Xbar<5.94) = 0.013516

(by using z-table)

Required probability = 0.013516

Question 2

Part a

We are given

µ = 509

σ = 112

We have to find P(X≥588) = 1 – P(X<588)

Z = (X - µ)/σ

Z = (588 – 509)/112

Z = 0.705357

P(Z< 0.705357) = P(X<588) = 0.759706

(by using z-table)

P(X≥588) = 1 – P(X<588)

P(X≥588) = 1 – 0.759706

P(X≥588) = 0.240294

Required probability = 0.240294

Part b

We are given

µ = 509

σ = 112

n = 18

We have to find P(Xbar≥588) = 1 – P(Xbar<588)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (588 – 509)/[112/sqrt(18)]

Z = 2.992577

P(Z< 2.992577) = P(Xbar<588) = 0.998617

(by using z-table)

P(Xbar≥588) = 1 – P(Xbar<588)

P(Xbar≥588) = 1 – 0.998617

P(Xbar≥588) = 0.001383

Required probability = 0.001383