The mean amount purchased by each customer at Churchill\'s Grocery Store is $18

Question

The mean amount purchased by each customer at Churchill's Grocery Store is $18 with a standard deviation of $5 The population is positively skewed for a sample of 53 customers, answer the following questions What is the likelihood the sample mean is at least $20? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) What is the likelihood the sample mean is greater than $17 but less than $20? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Within what limits will 98% of the sample means occur? (Round the final answer to 2 decimal places.)

Explanation / Answer

mean = 18

standard deviation = 5

sample size = 53

standard deviation of sample mean = 5/53 = 0.69

a) z for 20 = (20-18)/0.69 = 2.89

P(X<20) = 0.9981

P(X>20) = 1-0.9981 = 0.0019

b)z for 17 = (17-18)/0.69 = -1/0.69 = -1.45

P(X<17) = 0.0736

P(17<X<20) = 0.9981-0.0736 = 0.9245

c)let us find limits around mean 18 i.e 49% of values less than mean and 49% of values greather than mean

let the limits be (L,U)

P(L<x<mean) = P(mean<x<U) = 0.49

P(x<L) = 0.01 and P(x<U) =0.99

z for 0.01 = -2.33

z for 0.99 = 2.33

(L-18)/0.69 = -2.33

L = 18- 2.33*0.69 = 16.39

U = 18+2.33*0.69 = 19.60

so limits are (16.39,19.60)

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