7.33 It has been reported that households in the West spend an annual average of

Question

7.33 It has been reported that households in the West
spend an annual average of $6050 for groceries. Assume
a normal distribution with a standard deviation of
$1500.
a. What is the probability that a randomly selected Western
household spends more than $6350 for groceries?
b. How much money would a Western household have
to spend on groceries per year in order to be at the
99th percentile (i.e., only 1% of Western households
would spend more on groceries)?

Explanation / Answer

mean = 6050 SD = 1500 z = (6360-6050)/1500 z = 0.2066667 [use z table to get this on your own if needed] prob(z > 0.2066666) The z value for 99th percentile 2.3263 x = 2.3263*1500 + 6050 x = 9539.45

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