It has been reported that households in the West spend an
annual average of $6050 for groceries. Assume a normal distri-
bution 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?
NOTE: Please use attached Excel spreadsheet to show all computations. Please do not just give the answer but show fully exactly how you arrive at each answer. All computations should be complete and in agreement with all answers.
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Please see the attachments.
Let X be the annual expenditure of Western households for groceries. Given that X is normal with mean µ = 6050 and standard deviation = 1500.
We need P (X > 6350). Standardizing the variable X using
and from standard normal tables, we ...
The solution provides step by step method for the calculation of probability using the Z score. Formula for the calculation and Interpretations of the results are also included.