/*! 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 50 In a certain population, \(15 \%... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 50

In a certain population, \(15 \%\) of the people have Rh-negative blood. A blood bank serving this population receives 92 blood donors on a particular day. a. What is the probability that 10 or fewer are Rh-negative? b. What is the probability that 15 to 20 (inclusive) of the donors are Rh- negative? c. What is the probability that more than 80 of the donors are Rh-positive?

Short Answer

Expert verified
Answer: To calculate the probabilities, use the binomial probability formula and sum the probabilities for each relevant value of k: 1. P(X 鈮� 10) = 危 P(X = k) for k = 0 to 10 2. P(15 鈮� X 鈮� 20) = 危 P(X = k) for k = 15 to 20 3. P(X 鈮� 11) = 危 P(X = k) for k = 0 to 11 (Note: This is the same as having more than 80 Rh-positive donors)

Step by step solution

01

Calculate the probability that 10 or fewer are Rh-negative

We need to find the probability of having 0 to 10 Rh-negative donors. We will calculate the probabilities for each value of k (from 0 to 10) and sum them up: P(X 鈮� 10) = 危 P(X = k) for k = 0 to 10
02

Calculate the probability for 15 to 20 (inclusive) of the donors are Rh-negative

We need to find the probability of having 15 to 20 Rh-negative donors. We will calculate the probabilities for each value of k (from 15 to 20) and sum them up: P(15 鈮� X 鈮� 20) = 危 P(X = k) for k = 15 to 20
03

Calculate the probability that more than 80 of the donors are Rh-positive

We need to find the probability of having more than 80 Rh-positive donors, which means having fewer than 12 Rh-negative donors. We will calculate the probabilities for each value of k (from 0 to 11) and sum them up: P(X 鈮� 11) = 危 P(X = k) for k = 0 to 11 Note that P(X 鈮� 11) represents the probability of having 11 or fewer Rh-negative donors, which is the same as having more than 80 Rh-positive donors.

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 used-car dealership has found that the length of time before a major repair is required on the cars it sells is normally distributed with a mean equal to 10 months and a standard deviation of 3 months. If the dealer wants only \(5 \%\) of the cars to fail before the end of the guarantee period, for how many months should the cars be guaranteed?

Find the probability that \(z\) lies between \(z=-1.48\) and \(z=1.48\).

Philatelists (stamp collectors) often buy stamps at or near retail prices, but, when they sell, the price is considerably lower. For example, it may be reasonable to assume that (depending on the mix of a collection, condition, demand, economic conditions, etc.) a collection will sell at \(x \%\) of the retail price, where \(x\) is normally distributed with a mean equal to \(45 \%\) and a standard deviation of \(4.5 \% .\) If a philatelist has a collection to sell that has a retail value of \(\$ 30,000,\) what is the probability that the philatelist receives these amounts for the collection? a. More than \(\$ 15,000\) b. Less than \(\$ 15,000\) c. Less than \(\$ 12,000\)

A normal random variable \(x\) has an unknown mean \(\mu\) and standard deviation \(\sigma=2\). If the probability that \(x\) exceeds 7.5 is \(.8023,\) find \(\mu\).

One method of arriving at economic forecasts is to use a consensus approach. A forecast is obtained from each of a large number of analysts, and the average of these individual forecasts is the consensus forecast. Suppose the individual 2008 January prime interest rate forecasts of economic analysts are approximately normally distributed with the mean equal to \(8.5 \%\) and a standard deviation equal to \(0.2 \% .\) If a single analyst is randomly selected from among this group, what is the probability that the analyst's forecast of the prime rate will take on these values? a. Exceed \(8.75 \%\) b. Be less than \(8.375 \%\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Calculus

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.