Question 1162828: The mean amount purchased by each customer at Churchill's Grocery Store is $ 19 with a standard deviation of $9. The population is positively skewedFor a sample of 55 customers, answer the following questions a. What is the likelihood the sample mean is at least $20? (Round the value to 2 decimal places and the final answer to 4 decimal ploces.) Sample mean 0.2061 b. What is the likelihood the sample mean is greater than $16 but less than $20( Round the zwolue to 2 decimal places and the final answer to 4 declal places.) Sample mean 0.7871 c. Within what limits will 95% of the sample means occur(Round the final answers to 2 decimal places) Sample mean 16.62 and
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Using the central limit theorem...
z>(20-19)/9/sqrt(55)
z>0.82 and that is a probability of 0.2061
for 16, z=(16-19)/9/sqrt(55) or -2.47
probability of z between -2.47 and 0.82, the value for $20, is 0.7871
half-interval for 95% CI is 1.96*sigma/sqrt(n)=1.96*9/sqrt(55)=$2.38
the CI is ($16.62, $21.38), the $2.38's being subtracted from and added to the mean.
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