Question

The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions. What is the likelihood the sample mean is at least $25.00?

Answer 1

Answer: 0.0170

Explanation:

Given : The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00.

i.e.
`\mu=23.50`

`\sigma=5`

We assume the distribution of amounts purchased follows the normal distribution.

Sample size : n=50

Let
`\overline{x}` be the sample mean.

Formula :
`z=\frac{\overline{x}-\mu}{(\sigma)/(√(n))}`

Then, the probability that the sample mean is at least $25.00 will be :-

`P(\overline{x}\geq\25.00)=P(\frac{\overline{x}-\mu}{(\sigma)/(√(n))}\geq(25-23.50)/((5)/(√(50))))\\\\=P(z\geq2.12)\\\\=1-P(z<2.12)\\\\=1-0.9830=0.0170`

Hence, the likelihood the sample mean is at least $25.00= 0.0170