/*! 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 398 If \(Z\) is a standard normal va... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 398

If \(Z\) is a standard normal variable, use the table of standard normal probabilities to find: (a) \(\operatorname{Pr}(Z<0)\) (b) \(\operatorname{Pr}(-12.54)\).

Short Answer

Expert verified
In summary, using the table of standard normal probabilities, we find the following probabilities: (a) \(\operatorname{Pr}(Z<0) = 0.5\) (b) \(\operatorname{Pr}(-12.54) = 0.0056\)

Step by step solution

01

Standard Normal Probability Table

First, make sure you have a table of standard normal probabilities. These tables usually represent the area under the standard normal curve to the left of a given Z value. You can find these tables in your textbook or online (https://en.wikipedia.org/wiki/Standard_normal_table).
02

Understanding the Table

The table consists of rows representing tenths of the Z-score and columns representing hundredths of the Z-score. For example, to find the area under the curve to the left of Z=1.24, we would go down the rows to the row labeled 1.2, and then go across the columns to the column labeled 0.04, finding the value 0.8925 in the table.
03

Find the probability for Z

Since Z is a standard normal variable, the mean is 0. The table gives cumulative probabilities, that is, the area under the curve to the left of the given Z score. To find Pr(Z<0), we need to find the area under the curve to the left of 0. Since the standard normal distribution is symmetric about its mean, we know that Pr(Z<0) = Pr(Z>0) = 0.5.
04

Find the probability for -1

To find the probability Pr(-11)=1-Pr(Z<1)). Now, subtract the two areas: Pr(-1<Z<1) = 0.8413 - 0.1587 = 0.6826.
05

Find the probability for Z>2.54

To find Pr(Z>2.54), we need to find the area under the curve to the right of Z=2.54. Referring to the probability table, locate the value for Z=2.54, which is 0.9944. This value represents Pr(Z<2.54) or the area under the curve to the left of Z=2.54. Since we want the area to the right, we subtract from 1: Pr(Z>2.54) = 1 - 0.9944 = 0.0056. In summary, the probabilities are: a) Pr(Z<0) = 0.5 b) Pr(-12.54) = 0.0056

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

Measurements are made in \(\mathrm{mm}\) of the length of the trumpet in two varieties of daffodil ( \(\mathrm{A}\) and \(\mathrm{B}\) ). Variety A is crossed with variety \(\mathrm{B}\) and the same measurements are made on the \(\mathrm{F}_{1}\) and \(\mathrm{F} 2\) progeny (a) Calculate the mean, standard deviation, and standard error of the mean for each. (b) Determine whether the difference in the \(\mathrm{P}_{\mathrm{A}}\) and \(\mathrm{P}_{\mathrm{B}}\) varieties is statistically significant. (c) Determine whether the difference between the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) is statistically significant. (d) Assuming that each parental variety was homozygous and that the genes involved are additive in their effect, estimate how many gene pairs are segregating and assorting in the \(\mathrm{F}_{2}\) generation.

If \(\mathrm{X}\) has a normal distribution with mean 9 and standard deviation 3 , find \(\mathrm{P}(5<\mathrm{X}<11)\).

Consider that in humans red-green colorblindness is controlled by a recessive X-linked gene, c. Assume that a normal-visioned woman, whose father was colorblind, marries a normal-visioned man. What is the probability that their first child will be colorblind?

If a random variable \(\mathrm{X}\) is normally distributed with a mean of 118 and a standard deviation of 11 , what \(Z\) -scores correspond to raw scores of 115,134, and \(99 ?\)

Miniature poodles are thought to have a mean height of 12 inches and a standard deviation of \(1.8\) inches. If height is measured to the nearest inch, find the percentage of poodles having a height exceeding 14 inches.
See all solutions

Recommended explanations on Biology Textbooks

Heredity

Read Explanation

Biological Structures

Read Explanation

Reproduction

Read Explanation

Microbiology

Read Explanation

Biological Molecules

Read Explanation

Control of Gene Expression

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.