Question

QUESTION 4 The weigh of cans of salmon is randomly distributed with mean =6.05 and standard deviation = .18. The sample size is 36. What is the probability that the mean weight of the sample is more than 6.02? 4 decimals

QUESTION 5 The weigh of cans of salmon is randomly distributed with mean =6.05 and standard deviation = .18. The sample size is 36. What is the probability that the mean weight of the sample is between 5.99 and 6.07? 4 decimals

QUESTION 6 The weigh of cans of salmon is randomly distributed with mean =4 and standard deviation = 0.188. The sample size is 78. Find the Xbar value so that only 5% of the mean weight could be more than this Xbar value (case 4). 4 decimals

Answer 1

4)
Here, μ = 6.05, σ = 0.18/sqrt(36) = 0.03 and x = 6.02. We need to compute P(X >= 6.02). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (6.02 - 6.05)/0.03 = -1

Therefore,
P(X >= 6.02) = P(z <= (6.02 - 6.05)/0.03)
= P(z >= -1)
= 1 - 0.1587
= 0.8413

5)
z = (x - μ)/σ
z1 = (5.99 - 6.05)/0.03 = -2
z2 = (6.07 - 6.05)/0.03 = 0.67

Therefore, we get
P(5.99 <= X <= 6.07) = P((6.07 - 6.05)/0.03) <= z <= (6.07 - 6.05)/0.03)
= P(-2 <= z <= 0.67) = P(z <= 0.67) - P(z <= -2)
= 0.7486 - 0.0228
= 0.7258

6)
z-value = 1.645

x = 4 + 1.645*0.188/sqrt(78)
x = 4.04