/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q158SE Consider the discrete probabilit... [FREE SOLUTION] | 91影视

91影视

Consider the discrete probability distribution shown here.

x

10

12

18

20

p

.2

.3

.1

.4

a. Calculate,2 and .

b. What isP(x<15) ?

c. Calculate 2 .

d. What is the probability that xis in the interval 2 ?

Short Answer

Expert verified

(a) =15.4,2=18.44and=4.294.

(b) Px<15=0.5

(c) 2=(6.812,23.988)

(d) The required probability is 0.5.

Step by step solution

01

Given information

X is a random variable.

02

Identifying the type of random variable

a.

Mean=

=i=14xjPxi=100.2+120.3+180.1+200.4=2+3.6+1.8+8=15.4

Variance=2

2=i=14x-2Px=1015.420.2+12-15.420.3+18-15.420.1+20-15.420.4=29.160.2+11.560.3+6.760.1+21.160.4=5.832+3.468+0.676+8.464=18.44

Standard Deviation=

=18.44=4.294

Hence, =15.4,2=18.44 and =4.294 .

03

When  p(x<15)

b.

px<15=px=10+px=12=0.2+0.3=0.5

Hence, Px<15=0.5

04

Calculating the value when  μ±2σ

c.

The interval is

2=15.424.294=15.48.588=6.812,23.988

05

Calculating P (x) when μ±2σ

d.

P6.812x23.988=P0x17.176=1-Px17.176=1-Px=18+Px=20=1-0.1+0.4=1-0.5=0.5

Hence, the required probability is 0.5.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What are the treatments for a designed experiment with two factors, one qualitative with two levels (A and B) and one quantitative with five levels (50, 60, 70, 80, and 90)?

Redeeming tickets from textbook dispatches. Numerous companies now use textbook messaging on cell phones to sell their products. One way to do this is to shoot repairable reduction pasteboard (called an m- pasteboard) via a textbook. The redemption rate of m- tickets 鈥 the proportion of tickets redeemed 鈥 was the subject of a composition in the Journal of Marketing Research (October 2015). In a two-time study, over boardwalk shoppers shared by subscribing up to admit m-voucher. The experimenters were interested in comparing the redemption rates of m- tickets for different products in a sample of m- tickets for products vended at a milk-shake. Store, 79 were redeemed; in a sample of m- tickets for products vended at a donut store, 72 were redeemed.

a. Cipher the redemption rate for the sample of milk-shake m- tickets.

b. Cipher the redemption rate for the sample of donut m- tickets.

c. Give a point estimate for the difference between the actual redemption rates.

d. Form a 90 confidence interval for the difference between the actual redemption rates. Give a practical interpretation of the output.

e. Explain the meaning of the expression 鈥90 confident鈥 in your answer to part d.

f. Grounded on the interval, part d, is there a 鈥 statistically.鈥 A significant difference between the redemption rates? (Recall that a result is 鈥 statistically鈥 significant if there is substantiation to show that the true difference in proportions isn't 0.)

g. Assume the true difference between redemption rates must exceed.01 ( i.e., 1) for the experimenters to consider the difference 鈥 virtually鈥 significant. Based on the interval, part c, is there a 鈥渧irtually鈥 significant difference between the redemption rates?

To compare the means of two populations, independent random samples of 400 observations are selected from each population, with the following results:

Sample 1

Sample 2

x1=5,2751=150

x2=5,2402=200

a. Use a 95%confidence interval to estimate the difference between the population means (12). Interpret the confidence interval.

b. Test the null hypothesis H0:(12)=0versus the alternative hypothesis Ha:(12)0 . Give the significance level of the test and interpret the result.

c. Suppose the test in part b was conducted with the alternative hypothesis Ha:(12)0 . How would your answer to part b change?

d. Test the null hypothesis H0:(12)=25 versus Ha:(12)25. Give the significance level and interpret the result. Compare your answer with the test conducted in part b.

e. What assumptions are necessary to ensure the validity of the inferential procedures applied in parts a鈥揹?

Given the following values of x, s, and n, form a 90% confidence interval for2

a. x=21,s=2.5,n=50

b. x=1.3,s=0.02,n=15

c. x=167,s=31,n=22

d.x=9.4,s=1.5,n=5


A paired difference experiment yielded ndpairs of observations. In each case, what is the rejection region for testing H0:渭d>2?

a. nd=12,伪=.05

b.nd=24,伪=.10

c.nd=4,伪=.025

d.nd=80,伪=.01

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.