Pi. I don't know what the square root is doing there because it has no number under it.
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](https://tex.z-dn.net/?f=%5Cmu%3D23.50)
![\sigma=5](https://tex.z-dn.net/?f=%5Csigma%3D5)
We assume the distribution of amounts purchased follows the normal distribution.
Sample size : n=50
Let
be the sample mean.
Formula : ![z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B%5Coverline%7Bx%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
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](https://tex.z-dn.net/?f=P%28%5Coverline%7Bx%7D%5Cgeq%5C25.00%29%3DP%28%5Cdfrac%7B%5Coverline%7Bx%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%5Cgeq%5Cdfrac%7B25-23.50%7D%7B%5Cdfrac%7B5%7D%7B%5Csqrt%7B50%7D%7D%7D%29%5C%5C%5C%5C%3DP%28z%5Cgeq2.12%29%5C%5C%5C%5C%3D1-P%28z%3C2.12%29%5C%5C%5C%5C%3D1-0.9830%3D0.0170)
Hence, the likelihood the sample mean is at least $25.00= 0.0170
Mr. Webster is buying a carpet for an exercise room in his basement the room will an area of 230 ft
I believe it would be 25.7 miles. I just divide 822.4 by 32.