Simplify 14.2 + 6.5(p + 3q) – 4.05q
2 answers:
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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.
We assume the distribution of amounts purchased follows the normal distribution.
Sample size : n=50
Let be the sample mean.
Formula :
Then, the probability that the sample mean is at least $25.00 will be :-
Hence, the likelihood the sample mean is at least $25.00= 0.0170