The mean amount purchased by a typical customer at Churchill's Grocery Store is $23.50 with a standard deviation of $6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 52 customers, answer the following questions. |
a. |
What is the likelihood the sample mean is at least $25.50? (Round z value to 2 decimal places and final answer to 4 decimal places.) |
Probability |
b. |
What is the likelihood the sample mean is greater than $22.50 but less than $25.50? (Round z value to 2 decimal places and final answer to 4 decimal places.) |
Probability |
c. |
Within what limits will 95 percent of the sample means occur? (Round your answers to 2 decimal places.) |
Sample mean | and |
a)
To give that the likelihood the sample mean is at least $25.50
P(xbar >= 25.5)
= P((xbar - mean)/(s /vn) >(25.5 - 23.5)/(6 /sqrt(52)))
= P(Z > 2.40)
P(xbar >= 25.5) = 0.0082 [since from standard normal distribution table]
b)
To give that the likelihood the sample mean is greater than $22.50 but less than $25.50
P(22.5 < X < 25.5)
= P((22.5 - 23.5)/(6 /sqrt(52)) < Z < (25.5 - 23.5)/(6/sqrt(52)))
= P(-1.2 < Z < 2.40)
P(22.5 < X < 25.5) = 0.8767
c)
To determine the limits for 95% confidence interval
a = 0.05,
Z(0.025) = 1.96 (since from standard normal table)
Here 95% confidence interval is xbar +/- Z*s/vn
substitute values then we get limits
= 23.5 +/- 1.96*(6/sqrt(52)
Limits are = (21.87, 25.13)
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