/*! 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 4 Let \(X\) be \(N\left(\mu, \sigm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 4

Let \(X\) be \(N\left(\mu, \sigma^{2}\right)\) so that \(P(X<89)=0.90\) and \(P(X<94)=0.95\). Find \(\mu\) and \(\sigma^{2}\).

Short Answer

Expert verified
The mean (\mu) is approximately 72.42 and the variance (\sigma^{2}) is approximately 193.05.

Step by step solution

01

Find Z-values

Read from the standard normal distribution table for the z-values corresponding to the given probabilities. For \(P(X<89) = 0.90\), \(z_1\) approximately equals 1.28. For \(P(X<94) = 0.95\), \(z_2\) approximately equals 1.64.
02

Find Mu using the Z-value formulas

Use transformation formulas to solve for \(\mu\). This gives two equations: \(1.28 = (89 - \mu) / \sigma\) and \(1.64 = (94 - \mu) / \sigma\). Solve these two equations by subtraction to find the value of \(\mu\). By subtraction, \(0.36 = (5/\sigma)\). So, \(\mu = 94 - 1.64\sigma\)
03

Find Sigma using the found Mu

Substitute the value of \(\mu\) from Step 2 into any one of the two original equations. This gives an equation in \(\sigma\): \(1.28 = (89 - (94 - 1.64 \sigma)) / \sigma\), after simplifying becomes \(\sigma = 5 / 0.36\). Hence \(\sigma\) is approximately 13.889
04

Calculate Sigma Square

Calculate the variance \(\sigma^2\) by squaring the value of standard deviation \(\sigma\). Namely, \(\sigma^2 = (13.889)^2\)
05

Calculate Mu

Substitute the found value of \(\sigma\) into the equation \(\mu = 94 - 1.64\sigma\) to solve for \(\mu\). This gives \(\mu\) approximately equals 72.42

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

. Let \(Y\) have a truncated distribution with pdf \(g(y)=\phi(y) /[\Phi(b)-\Phi(a)]\), for \(a

Let \(X_{1}, X_{2}\), and \(X_{3}\) be iid random variables, each with pdf \(f(x)=e^{-x}\), \(0y)=1-P\left(X_{i}>y, i=1,2,3\right)\).

Let \(f(x)\) and \(F(x)\) be the pdf and the cdf of a distribution of the continuous type such that \(f^{\prime}(x)\) exists for all \(x\). Let the mean of the truncated distribution that has pdf \(g(y)=f(y) / F(b),-\infty

Let the number of chocolate drops in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate drops to be greater than \(0.99 .\) Find the smallest value of the mean that the distribution can take.

. Let \(X\) and \(Y\) have a bivariate normal distribution with parameters \(\mu_{1}=\) \(\mu_{2}=0, \sigma_{1}^{2}=\sigma_{2}^{2}=1\), and correlation coefficient \(\rho .\) Find the distribution of the random variable \(Z=a X+b Y\) in which \(a\) and \(b\) are nonzero constants.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Mechanics Maths

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.