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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 $48 and the estimated standard deviation is about $7.

(a) Consider a random sample of n = 60 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 μx = 48 and standard error σx = $0.12. The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $7. The sampling distribution of x is not normal. The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $0.90.


Is it necessary to make any assumption about the x distribution? Explain your answer.

It is necessary to assume that x has an approximately normal distribution. It is not necessary to make any assumption about the x distribution because μ is large. It is necessary to assume that x has a large distribution. It is not necessary to make any assumption about the x distribution because n is large.


(b) What is the probability that x is between $46 and $50? (Round your answer to four decimal places.)
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(c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $46 and $50? (Round your answer to four decimal places.)

Respuesta :

cchilabert

Answer:

Step-by-step explanation:

Hello!

X: amount spent in a supermarket impulse buying in a 10 min unplanned shopping interval by one customer.

It is known that the mean of this variable is μ= $48 and its standard deviation is δ=$7

a.

The Central limit theorem states that if there is a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

X[bar]≈N(μ;σ²/n)

The mean of the sampling distribution is μ= $48

The standard deviation of the sampling distribution is σ/√n= $7/√60= $0.90

It is not necessary to make any assumption about the distribution of X since n=60 is considered large enough, you can directly approximate the sampling distribution to normal regardless of the distribution of X.

b.

To calculate this probability you have to use the approximation of the sampling distribution:

Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

μ= $48

σ/√n= $0.90

P(46≤X[bar]≤50)= P(X[bar]≤50) - P(X[bar]≤46)

P(Z≤(50-48)/0.90) - P(Z≤(46-48)/0.90)

P(Z≤2.22) - P(Z≤-2.22)= 0.987 - 0.013= 0.974

c.

If we assume that X has an approximately normal distribution, then you will use it's a mean and standard deviation to reach the asked probability.

Z= (X-μ)/δ≈N(0;1)

μ= $48

δ= $7

P(46≤X≤50)= P(X≤50) - P(X≤46)

P(Z≤(50-48)/7) - P(Z≤(46-48)/7)

P(Z≤0.29) - P(Z≤-0.29)= 0.61409 - 0.38591= 0.22818≅ 0.2282

I hope it helps!

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