Question

-/2 POINTS MY NOTES ASK YOUR TEACHER Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a certain article, the mean of the x distribution is about $42 and the estimated standard deviation is about $7. (a) Consider a random sample of n = 80 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the x distribution? The sampling distribution of x is approximately normal with mean 4x = 42 and standard deviation Oy = $0.78. The sampling distribution of x is approximately normal with mean ux = 42 and standard deviation Ox = $7. The sampling distribution of x is not normal. The sampling distribution of x is approximately normal with mean ux = 42 and standard deviation Ox = $0.09. Is it necessary to make any assumption about the shape of the x distribution? Explain your answer. It is necessary to assume since the population size was not given. It is not necessary to make any assumption about the x distribution since the sample size, n, is large. It is not necessary to make any assumption about the x distribution because both u and o are given and they are very large. It is necessary to assume that the shape of the x distribution is approximately Normal even though it is not stated. (b) What is the probability that is between $40 and $44? (Round your answer to four decimal places.)

Answer 1

Question (a)

Given Population mean $\mu$ = 42

Population Standard deviation $\sigma$ = 7

Given Sample size n = 80

From the central limit theorem we can say that the probability distribution or sampling distribution of $\bar{x}$ is approximately normal with mean equal to $\mu$ (population mean) and Standard deviation equal to $\sigma$ / $\sqrt{}$ n where $\sigma$ is the population standard deviation and n is the sample size

Here the probability distribution or sampling distribution of $\bar{x}$ is not normal, it is approximately normal

Mean of the probability distribution of $\bar{x}$ , $\bar{\mu }$ x= 42 (Population Mean)

Standard deviation of the probability distribution of $\bar{x}$ , $\bar{\sigma }$ x = 7 / $\sqrt{}$ 80

= 7 / 8.9443

= 0.7826

= 0.78 rounded to 2 decimals

So the Answer is Option A

The sampling distribution of $\bar{x}$ is approximately normal with mean $\bar{\mu }$ x= 42 and Standard deviation $\bar{\sigma }$ x = 0.78

Question (b)

It is not necessary to make any assumption about the x distribution since the sample size n,is large

And the sample size increaes, according to central limit theorem, the sampling distribution converges on to a approximate nromal distribution but not completely normal distribution

So Asnwer is B

Question (c)

Probability that $\bar{x}$ lies between 40 and 44

We need to calculate the respective z-scores for 40 and 33 and then the area to the left of these z-scores by using the Z-tables. Then we subtract one area from the other to arrive at Probability that $\bar{x}$ lies between 40 and 44

Here $\bar{\mu }$ x= 42

$\bar{\sigma }$x = 0.78

We should consider the mean and standard deviation of sampling distribution of $\bar{x}$ here since we are trying to find the Probability that $\bar{x}$ lies between 40 and 44

Z-score = (X - $\bar{\mu }$ x ) / $\bar{\sigma }$ x

For X = 40

Z-score = (40 - 42) / 0.7826

Z-score = -2 / 0.7826

Z-score = -2.5556

We can find the area to the left of z-score using the below z-tables. Here we should use the negative z-table and also the z-scores should be rounded to 2 decimals. So here z-score = -2.56

The area to left of z-score of -2.56 from the negative z-table attached below is 0.00539

The exact vale of area to the left of z-score -2.5556 can be calculated from the online calculators. The area to the left of z-score -2.5556 is 0.0053002, so rounding to 4 decimals it is 0.0053

So P($\bar{x}$ < 40) = 0.0053

For X = 44

Z-score = (44 - 42) / 0.7826

Z-score = 2 / 0.7826

Z-score = 2.5556

We can find the area to the left of z-score using the below z-tables. Here we should use the positive z-table and also the z-scores should be rounded to 2 decimals. So here z-score = 2.56

The area to left of z-score of 2.56 from the positive z-table attached below is 0.99461

The exact vale of area to the left of z-score 2.5556 can be calculated from the online calculators. The area to the left of z-score 2.5556 is 0.99469, so rounding to 4 decimals it is 0.9947

So P($\bar{x}$ < 44) = 0.9947

P(42 < $\bar{x}$ < 44) = P($\bar{x}$ < 44) - P($\bar{x}$ < 40)

= 0.9947 - 0.0053

= 0.9894

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