/*! 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} Problem 55 Find the following probabilities... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 55

Find the following probabilities for the standard normal random variable: a. \(P(.3

Short Answer

Expert verified
Answer: The probabilities for the given standard normal random variable are: a. \(P(.3<z<1.56) = 0.3227\) b. \(P(-.2<z<.2) = 0.1586\)

Step by step solution

01

Use the Z-table for the first interval (\(P(.3

To find the probabilities for the first interval, refer to the Z-table to look up the probabilities for \(z = .3\) and \(z = 1.56\). Based on the Z-table, we get: - \(P(z \le .3) = 0.6179\) - \(P(z \le 1.56) = 0.9406\)
02

Calculate the probability for the first interval

Now, we will subtract the probability for the lower limit from the probability for the upper limit to find the probability between these points: \(P(.3 < z < 1.56) = P(z \le 1.56) - P(z \le .3)\) \(P(.3 < z < 1.56) = 0.9406 - 0.6179 = 0.3227\) The probability for the first interval is 0.3227.
03

Use the Z-table for the second interval (\(P(-.2

Repeat the process for the second interval. Refer to the Z-table to look up the probabilities for \(z = -.2\) and \(z = .2\). Based on the Z-table, we get: - \(P(z \le -.2) = 0.4207\) - \(P(z \le .2) = 0.5793\)
04

Calculate the probability for the second interval

Again, subtract the probability for the lower limit from the probability for the upper limit to find the probability between these points: \(P(-.2 < z < .2) = P(z \le .2) - P(z \le -.2)\) \(P(-.2 < z < .2) = 0.5793 - 0.4207 = 0.1586\) The probability for the second interval is 0.1586. In summary, the probabilities for the given standard normal random variable are: a. \(P(.3<z<1.56) = 0.3227\) b. \(P(-.2<z<.2) = 0.1586\)

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

a. Find a \(z_{0}\) such that \(P\left(-z_{0} \leq z \leq z_{0}\right)=.95\). b. Find a \(z_{0}\) such that \(P\left(-z_{0} \leq z \leq z_{0}\right)=.98\).

It is known that \(30 \%\) of all calls coming into a telephone exchange are long-distance calls. If 200 calls come into the exchange, what is the probability that at least 50 will be long-distance calls?

The life span of oil-drilling bits depends on the types of rock and soil that the drill encounters, but it is estimated that the mean length of life is 75 hours. Suppose an oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 75 hours and a standard deviation equal to 12 hours. a. What proportion of the company's drill bits will fail before 60 hours of use? b. What proportion will last at least 60 hours? c. What proportion will have to be replaced after more than 90 hours of use?

A researcher notes that senior corporation executives are not very accurate forecasters of their own annual earnings. He states that his studies of a large number of company executive forecasts "showed that the average estimate missed the mark by \(15 \%\)." a. Suppose the distribution of these forecast errors has a mean of \(15 \%\) and a standard deviation of \(10 \%\). Is it likely that the distribution of forecast errors is approximately normal? b. Suppose the probability is .5 that a corporate executive's forecast error exceeds \(15 \% .\) If you were to sample the forecasts of 100 corporate executives, what is the probability that more than 60 would be in error by more than \(15 \% ?\)

Find the normal approximation to \(P(355 \leq x \leq 360)\) for a binomial probability distribution with \(n=400\) and \(p=.9\).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation
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.