Question

A small shop has investigated its customers to determine how
much money they spend in shop. The study revealed that the spending
distribution is approximately normally distributed with a mean of
OR 4.11 and standard deviation of OR 1.37.

(i) What percentage of customers will spend less than OR 3.1
in the shop.

(ii) What spending amount corresponds to the top 87th
percentile?

(iii) Suppose owner takes a random sample of 25 customers and
record their spending. What is the probability that his sample
average will spend less than OR 3.1 in the shop?

(iv) Compare the results compared in (i) and (iii)
above.

Answer #1

I used R software to solve this question.

R codes and output:

pnorm(3.1,4.11,1.37)

[1] 0.2304924

qnorm(0.13,11,1.37)

[1] 9.456844

s=1.37/sqrt(25)

pnorm(3.1,4.11,s)

[1] 0.0001138444

Que.1

P(X < 3.1 ) = 0.2305

Hence percentage of customers will spend less than OR 3.1 in the shop is 23.05%

Que.2

Top 87 percentile means area less than that point is 100 - 87 = 13

Hence

Hence spending amount corresponds to top the top 87th percentile is 9.46

Que.3

Means of sampling distribution of mean is 3.1 and standard deviation is

P(X < 3.1 ) = 0.00011

Que.4

Probability in part (iii) is very less as compared to probability in part (i).

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