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Chapter 8: Q132E (page 452)

4.132 Suppose xis a random variable best described by a uniform

probability distribution with c= 3 and d= 7.

a. Find f(x)

b. Find the mean and standard deviation of x.

c. FindP(μ-σ≤x≤μ+σ)

Short Answer

Expert verified

a. The probability density function is

f(x)=0.253≤x≤70;otherwise

b. The mean is 5 and standard deviation is 1.1547.

c.P(μ-σ≤x≤μ+σ)

Step by step solution

01

Given Information

Here, x is a uniform random variable with parameters c=3 and d=7.

02

Finding the pdf of x

a.

The probability density function random variable x is given by

f(x)=1d-c;c<x<d

Here, c=3 and d=7.

So, the pdf of x is:

f(x)=17-3=14=0.25

Thus, f (x) = 0.25 ; 3 < x < 7

03

Finding the mean and standard deviation of x.

b.

The mean of the random variable x is given by,

μ=c+d2=3+72=102=5

The standard deviation of x is given by,

σ=d-c12=7-312=423=23=1.1547

Thus, the mean μ=5and standard deviation σ=1.1547.

04

Finding the P(μ-σ≤x≤μ+σ)

c.P(μ-σ≤x≤μ+σ)=∫μ-σμ+σf(x)dx=∫μ-σμ+σ0.25dx=0.25∫μ-σμ+σdx=0.25xμ-σμ+σ=0.25μ+σ-μ+σ=0.25×2σ=0.50×23=0.5774

So, the required probability is 0.5774.

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Most popular questions from this chapter

In a random sample of 250 people from a city, 148 of them favor apples over other fruits.

a. Use a 90% confidence interval to estimate the true proportion p of people in the population who favor apples over other fruits.

b. How large a sample would be needed to estimate p to be within .15 with 90% confidence?

The gender diversity of a large corporation’s board of directors was studied in Accounting & Finance (December 2015). In particular, the researchers wanted to know whether firms with a nominating committee would appoint more female directors than firms without a nominating committee. One of the key variables measured at each corporation was the percentage of female board directors. In a sample of 491firms with a nominating committee, the mean percentage was 7.5%; in an independent sample of 501firms without a nominating committee, the mean percentage was role="math" localid="1652702402701" 4.3% .

a. To answer the research question, the researchers compared the mean percentage of female board directors at firms with a nominating committee with the corresponding percentage at firms without a nominating committee using an independent samples test. Set up the null and alternative hypotheses for this test.

b. The test statistic was reported as z=5.1 with a corresponding p-value of 0.0001. Interpret this result if α=0.05.

c. Do the population percentages for each type of firm need to be normally distributed for the inference, part b, to be valid? Why or why not?

d. To assess the practical significance of the test, part b, construct a 95% confidence interval for the difference between the true mean percentages at firms with and without a nominating committee. Interpret the result.

Predicting software blights. Relate to the Pledge Software Engineering Repository data on 498 modules of software law written in “C” language for a NASA spacecraft instrument, saved in the train. (See Exercise 3.132, p. 209). Recall that the software law in each module was estimated for blights; 49 were classified as “true” (i.e., the module has imperfect law), and 449 were classified as “false” (i.e., the module has corrected law). Consider these to be Arbitrary independent samples of software law modules. Experimenters prognosticated the disfigurement status of each module using the simple algorithm, “If the number of lines of law in the module exceeds 50, prognosticate the module to have a disfigurement.” The accompanying SPSS printout shows the number of modules in each of the two samples that were prognosticated to have blights (PRED_LOC = “yes”) and prognosticated to have no blights (PRED_LOC = “no”). Now, define the delicacy rate of the algorithm as the proportion of modules. That was rightly prognosticated. Compare the delicacy rate of the algorithm when applied to modules with imperfect law with the delicacy rate of the algorithm when applied to modules with correct law. Use a 99-confidence interval.

DEFECT*PRED_LOC crosstabulation


PRED_LOC
 total
noyes

DEFECT False

True

total

440

29

429

49

20

69

449

49

498

Question: Independent random samples from approximately normal populations produced the results shown below.

Sample 1

Sample 2

52 33 42 4441 50 44 5145 38 37 4044 50 43

52 43 47 5662 53 61 5056 52 53 6050 48 60 55

a. Do the data provide sufficient evidence to conclude that (μ1-μ2)>10? Test usingα=0.1.

b. Construct a confidence interval for (μ1-μ2). Interpret your result.

Two populations are described in each of the following cases. In which cases would it be appropriate to apply the small-sample t-test to investigate the difference between the population means?

a.Population 1: Normal distribution with variance σ12. Population 2: Skewed to the right with varianceσ22=σ12.

b. Population 1: Normal distribution with variance σ12. Population 2: Normal distribution with variance σ22≠σ12.

c. Population 1: Skewed to the left with variance σ12. Population 2: Skewed to the left with varianceσ22=σ12.

d. Population 1: Normal distribution with varianceσ12 . Population 2: Normal distribution with varianceσ22=σ12 .

e. Population 1: Uniform distribution with varianceσ12 . Population 2: Uniform distribution with variance σ22=σ12.
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