/*! 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} Q. 3 T2.3. Rainwater was collected in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q. 3 (page 138)

T2.3. Rainwater was collected in water collectors at 30different sites near an industrial complex, and the amount of acidity (pH level) was measured. The mean and standard deviation of the values are 4.60and 1.10, respectively. When the pH meter was recalibrated back at the laboratory, it was found to be in error. The error can be corrected by adding 0.1pHunits to all of the values and then multiplying the result by 1.2. The mean and standard deviation of the corrected pH measurements are

(a)5.64,1.44

(b)5.64,1.32

(c)5.40,1.44

(d)5.40,1.32

(e)5.64,1.20

Short Answer

Expert verified

The mean and standard deviation of the corrected pH measurements are(b)5.64,1.32.

Step by step solution

01

Given Information

Given:

Number of sites=30

Mean value=4.60

Standard deviation value=1.10

Adding pH=0.1pH

02

Explanation

Given:

渭X=4.60

蟽X=1.10

Add 0.1to every value and then multiply by :

1.2(X+0.1)=1.2X+0.12

Properties mean and standard deviation:

渭ax+b=a渭X+b

蟽aX+b=a蟽X

Find the mean and standard deviation for 1.2X+0.12:

localid="1649415494484" 渭1.2X+0.12=1.2渭X+0.12=1.2(4.60)+0.12=5.64

localid="1649415506162" 蟽1.2X+0.12=1.2蟽X=1.2(1.10)=1.32.

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

Biking accidents Accidents on a level, 3-mile bike path occur uniformly along the length of the path. The figure below displays the density curve that

describes the uniform distribution of accidents.

(a) Explain why this curve satisfies the two requirements for a density curve.

(b) The proportion of accidents that occur in the first mile of the path is the area under the density curve between 0 miles and 1 mile. What is this area?

(c) Sue鈥檚 property adjoins the bike path between the 0.8-mile mark and the 1.1-mile mark. What proportion of accidents happens in front of Sue鈥檚 property?

Explain.

Density curves Sketch a density curve that might describe a distribution that has a single peak and is skewed to the left.

Follow the method shown in the examples to answer each of the following questions. Use your calculator or the Normal Curve applet to check your answers.

1. Cholesterol levels above 240mg/dlmay require medical attention. What percent of 14-year-old boys have more than 240mg/dlof cholesterol?

Weights aren鈥檛 Normal The heights of people of the same gender and similar ages follow Normal distributions reasonably closely. Weights, on the other hand, are not Normally distributed. The weights of women aged 20 to 29 have mean 141.7 pounds and median 133.2 pounds. The first and third quartiles are 118.3 pounds and 157.3 pounds. What can you say about the shape of the weight distribution? Why?

T2.11. As part of the President's Challenge, students can attempt to earn the Presidential Physical Fitness Award or the National Physical Fitness Award by meeting qualifying standards in five events: curl-ups, shuttle run, sit and reach, one-mile run, and pull-ups. The qualifying standards are based on the 1985School Population Fitness Survey. For the Presidential award, the standard for each event is the 85thpercentile of the results for a specific age group and gender among students who participated in the 1985survey. For the National award, the standard is the 50th percentile. To win either award, a student must meet the qualifying standard for all five events.
Jane, who is 9years old, did 40curl-ups in one minute. Matt, who is 12years old, also did 40curl-ups in one minute. The qualifying standard for the Presidential award is 39curl-ups for Jane and 50curl-ups for Matt. For the National award, the standards are 30and 40, respectively.
(a) Compare Jane's and Matt's performances using percentiles. Explain in language simple enough for someone who knows little statistics to understand.
(b) Who has the higher standardized value (z-score), Jane or Matt? Justify your answer.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

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.