Question

proportion (P) with sample size 100. Question 6: Compute the standard deviation of P in Question 5 above. Question 7: Randomly choose 100 customers. Find the probability that less than 40 of them spent less than ¥800. Walmart conducted a similar study on customers in Guangzhou (GZ) the same year. It was found that the amounts those customers spent were normally distributed with an average of ¥1,950 and a standard deviation of Y320. Question 8: A random sample of 64 data was selected from this population. What is the probability that the mean of the sample in Question 3 is less than the mean of this sample?

Walmart is an American multinational retail corporation that operates a chain of large

department stores. In 2017, they conducted a study on the amounts customers spent at their

stores in Shenzhen (SZ) on the day of New Year’s Eve. It was found that the amounts were

normally distributed, with an average of ¥1,800 and a standard deviation of ¥240.

Question 1:

Consider sampling with sample size 25 on the above population. Compute the mean of the

sampling distribution of the mean ().

Question 2:

Compute the standard deviation of in Question 1 above.

Question 3:

Suppose 25 customers were randomly selected. What is the probability that they spent more than

¥1,850 on average?

Question 4:

Would the calculation you performed in Question 3 still be valid if the amounts were NOT

normally distributed? Why?

In the study above, it was also estimated that 32% of the SZ customers spent less than ¥800.

Question 5:

Compute the mean of the sampling distribution of proportion () with sample size 100.

Question 6:

Compute the standard deviation of in Question 5 above.

Question 7:

Randomly choose 100 customers. Find the probability that less than 40 of them spent less than

¥800.

Walmart conducted a similar study on customers in Guangzhou (GZ) the same year. It was found

that the amounts those customers spent were normally distributed with an average of ¥1,950 and

a standard deviation of ¥320.

Question 8:

A random sample of 64 data was selected from this population. What is the probability that the

Answer 1

Solution5:

standard deviation=sqrt(p*(1-p)/n)

=sqrt(0.32*(1-0.32)/100)

= 0.04664762

Solution-6:

p=0.32

p^=40/100=0.40

P(p^<0.32)

z=p^-p/sqrt(p*(1-p)/n)

P(Z<0.40-0.32/0.04664762)

=P(Z< 1.714986)

=P(Z<1.71)

=0.9564