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).
Get Answers For Free
Most questions answered within 1 hours.