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Natali5045456 [20]
3 years ago
7

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

me 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?
Mathematics
1 answer:
m_a_m_a [10]3 years ago
5 0

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

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

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