SOLUTION: The mean amount purchased by each customer at Churchill’s Grocery Store is $35 with a standard deviation of $9. The population is positively skewed. For a sample of 42 customers,
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Question 1152710: The mean amount purchased by each customer at Churchill’s Grocery Store is $35 with a standard deviation of $9. The population is positively skewed. For a sample of 42 customers, answer the following questions:
a. What is the likelihood the sample mean is at least $39? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)
Sample mean
b. What is the likelihood the sample mean is greater than $33 but less than $39? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)
Sample mean
c. Within what limits will 98% of the sample means occur? (Round the final answers to 2 decimal places.)
You can put this solution on YOUR website! Assuming I can use normality, since otherwise the problem can't be done without knowing the skew.
z=(xbar-35)/9/sqrt(42)
at least 39 for sample mean
z>(4*sqrt(42)/9)= 2.88
that probability is 0.0020
between 33 and 39
z is -2*sqrt42/9 for $33, or -1.44
probability z is between -1.44 and 2.88 is 0.9231
half-interval for 98%CI is z=2.33 and the interval is z sigma/sqrt(n)=3.24
98% sample means will be ($31.76, $38.24)
