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: 0.0170Step-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. [tex]\mu=23.50[/tex][tex]\sigma=5[/tex]We assume the distribution of amounts purchased follows the normal distribution.Sample size : n=50Let [tex]\overline{x}[/tex] be the sample mean.Formula : [tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]Then, the probability that the sample mean is at least $25.00 will be :-[tex]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[/tex]Hence, the likelihood the sample mean is at least $25.00= 0.0170