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

Step-by-step 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 : $z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

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

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

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