Question

# (8) Let x be a random variable that represents the level of glucose in the blood...

(8) Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12 hour fast. Assume that for people under 50 years old, x has a distribution that is approximately normal, with mean μ = 54 and estimated standard deviation σ = 11. A test result x < 40 is an indication of severe excess insulin, and medication is usually prescribed.

(a) What is the probability that, on a single test, x < 40? (Round your answer to four decimal places.)

(b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? Hint: See Theorem 6.1.

What is the probability that x < 40? (Round your answer to four decimal places.)

(c) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to four decimal places.)

(d) Repeat part (b) for n = 5 tests taken a week apart. (Round your answer to four decimal places.)

Explain what this might imply if you were a doctor or a nurse.

(9) 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 \$27 and the estimated standard deviation is about \$9.

(a) Consider a random sample of n = 100 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?

(b) What is the probability that x is between \$25 and \$29? (Round your answer to four decimal places.)

(c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between \$25 and \$29? (Round your answer to four decimal places.)

(d) In part (b), we used x, the average amount spent, computed for 100 customers. In part (c), we used x, the amount spent by only one customer. The answers to parts (b) and (c) are very different. Why would this happen?

10) A European growth mutual fund specializes in stocks from the British Isles, continental Europe, and Scandinavia. The fund has over 325 stocks. Let x be a random variable that represents the monthly percentage return for this fund. Suppose x has mean μ = 1.4% and standard deviation σ = 1.1%.

(a) Let's consider the monthly return of the stocks in the fund to be a sample from the population of monthly returns of all European stocks. Is it reasonable to assume that x (the average monthly return on the 325 stocks in the fund) has a distribution that is approximately normal? Explain.

(Blank)  x is a mean of a sample of n = 325 stocks. By the(Blank)  the x distribution( Blank) approximately normal?

(b) After 9 months, what is the probability that the average monthly percentage return x will be between 1% and 2%? (Round your answer to four decimal places.)(c)After 18 months, what is the probability that the average monthly percentage return x will be between 1% and 2%? (Round your answer to four decimal places.

(d) Compare your answers to parts (b) and (c). Did the probability increase as n (number of months) increased? Why would this happen?

(e) If after 18 months the average monthly percentage return x is more than 2%, would that tend to shake your confidence in the statement that μ = 1.4%? If this happened, do you think the European stock market might be heating up? (Round your answer to four decimal places.)  P(x > 2%)

Explain:

8)

As the given data has normal distribution, we can use standard normal z table to estimate the answers

Given mean = 54

S.d = 11

Z = (x-mean)/s.d

A)

We need to find

P(x<40)

Z = (40-54)/11

Z = -1.27

From z table, p(z<-1.27) = 0.1020

B)

In case of a sample

Z = (x-mean)/(s.d/√n)

N = 2 here

Z = (40-54)/(11/√2)

Z = -1.8

From z table, p(z<-1.8) = 0.0359

C)

N = 3

Z = (40-54)/(11/√3)

Z = -2.2

From z table, p(z<-2.2) = 0.0139

D)

N = 5

Z = (40-54)/(11/√5)

Z = -2.85

From z table, p(z<-2.85) = 0.0022

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