Question

The mean amount purchased by each customer at Churchill’s Grocery Store is $29 with a standard deviation of $8. The population is positively skewed. For a sample of 58 customers, answer the following questions: a. What is the likelihood the sample mean is at least $31? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Sample mean b. What is the likelihood the sample mean is greater than $26 but less than $31? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Sample mean c. Within what limits will 98% of the sample means occur? (Round the final answers to 2 decimal places.) Sample mean and

Answer 1

μ =29

σ = 8

n = 58

A)

P(X >= 31)

Z = X - μ /(σ /√n)

= 31 - 29/(8/√58)

= 1.90

P( z >= 1.90)

=0.5 - 0.4713

= 0.0287

B)

P(26 < X < 31)

= P[26-29/(8/√58) < z < 31-29/(8/√58) ]

= P(-2.85 < z < 1.90)

= 0.4978 + 0.4713 [standard normal distribution table ]

= 0.9691

C)

98% of the sample means 49% less than mean and 49% more than mean value

From standard normal distribution table

At p = 0.49 , z = +/- 2.325

Z1 = X1 - μ / ( σ / √n)

-2.325 = X1 - 29/(8 /√58)

X1 = 26.5577

Z2 = X2 - μ / ( σ / √n)

2.325 = X2 - 29 /(8/√58)

X2 = 31.4423

Confidence interval (26.5577 , 31.4423)