BEGINNING STATISTICS 2E TEXTBOOK+BEGIN
BEGINNING STATISTICS 2E TEXTBOOK+BEGIN
2nd Edition
ISBN: 9781642770582
Author: WARREN DENLEY
Publisher: HAWKES LRN
Question
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Chapter 3.1, Problem 14E
To determine

To find:

The mean, median, and mode for the given data set.

Expert Solution & Answer
Check Mark

Answer to Problem 14E

Solution:

The mean temperature is 87.5°F, the median temperature is 88.5°F, and the data is multimodal with modal temperatures 85°F,88°F, 89°F, 90°F, 91°F, and92°F.

Explanation of Solution

Given information:

The given data set is,

High Temperatures for Cities in the Southeast

Stem Leaves
7
7
8
8
9
9

7 9
2 3 4
5 5 7 8 8 9 9
0 0 1 1 2 2 3
5

Key:7 7=77°F

Procedure:

A stem leaf plot is combination of two columns. Last significant digit of each data should be placed in the column of leaves and others digit in the column of stem. Each stem should be listed in numerical order and each leaf should be placed next to its stem.

Definition

Mean is the average of the given data set.

Median is the mid value of ordered data set.

Mode is the most frequently occurred value in the ordered data set.

If all the data values occur only once, then the data set has no mode. If only one value occurs most often, then the data set is unimodal. If exactly two values occur equally often, then the data set is bimodal. If more than two values occur equally often, then the data set is multimodal.

Following is the set of data corresponding to the given stem and leaf plot.

77°,79°,82°,83°,84°,85°,85°,87°,88°,88°,89°,89°,90°,90°,91°,91°,92°,92°,93°,95°.

Formula used:

The formula to calculate the mean of set of data is,

Mean(x¯)=x1+x2+...+xnn =i=1nxin

Where, xi is the ith data value and n is the number of data values in the data set.

Calculation:

Substitute 77,79,...,95 for x1,x2,x3,...,x20 respectively in the formula.

Here number of data in the given data set is 20, substitute 20 for n in the formula.

Mean(x¯)=x1+x2+...+xnn =77+79+82+83+84+85+85+87+88+88+89+89+90+90+91+91+92+92+93+9520 =175020 =87.5

Mean temperature is 87.5°F.

Formula used:

The formula to calculate the median of set of ordered data is,

Median={(n+12)thvalue,n=odd12[(n2)thvalue+(n+12)thvalue],n=even

n is the number of data values in the data set.

Calculation:

Here given set of data is,

77°,79°,82°,83°,84°,85°,85°,87°,88°,88°,89°,89°,90°,90°,91°,91°,92°,92°,93°,95°

The data is already an ordered data.

Number of values in data set is 20, it is even.

Substitute 20 for n in the formula for even number of data set.

Median=12[(n2)thvalue+(n2+1)thvalue]=12[(202)thvalue+(202+1)thvalue]=12[10thvalue+11thvalue]=12[88+89]

Simplify further,

Median=1772=88.5

The median temperature is 88.5°F.

The most frequently occurred data is the mode.

Here, each of the data 85, 88, 89, 90, 91, 92 occur 2 times.

Modes of the given data set are 85°F,88°F, 89°F, 90°F, 91°F, and92°F.

6 data occur most often, so the data set is multimodal.

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Chapter 3 Solutions

BEGINNING STATISTICS 2E TEXTBOOK+BEGIN

Ch. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.1 - Prob. 18ECh. 3.1 - Prob. 19ECh. 3.1 - Prob. 20ECh. 3.1 - Prob. 21ECh. 3.1 - Prob. 22ECh. 3.1 - Prob. 23ECh. 3.1 - Prob. 24ECh. 3.1 - Prob. 25ECh. 3.1 - Prob. 26ECh. 3.1 - Prob. 27ECh. 3.1 - Prob. 28ECh. 3.1 - Prob. 29ECh. 3.1 - Prob. 30ECh. 3.1 - Prob. 31ECh. 3.1 - Prob. 32ECh. 3.1 - Prob. 33ECh. 3.1 - Prob. 34ECh. 3.2 - Prob. 1ECh. 3.2 - Prob. 2ECh. 3.2 - Prob. 3ECh. 3.2 - Prob. 4ECh. 3.2 - Prob. 5ECh. 3.2 - Prob. 6ECh. 3.2 - Prob. 7ECh. 3.2 - Prob. 8ECh. 3.2 - Prob. 9ECh. 3.2 - Prob. 10ECh. 3.2 - Prob. 11ECh. 3.2 - Prob. 12ECh. 3.2 - Prob. 13ECh. 3.2 - Prob. 14ECh. 3.2 - Prob. 15ECh. 3.2 - Prob. 16ECh. 3.2 - Prob. 17ECh. 3.2 - Prob. 18ECh. 3.2 - Prob. 19ECh. 3.2 - Prob. 20ECh. 3.2 - Prob. 21ECh. 3.2 - Prob. 22ECh. 3.2 - Prob. 23ECh. 3.2 - Prob. 24ECh. 3.2 - Prob. 25ECh. 3.2 - Prob. 26ECh. 3.2 - Prob. 27ECh. 3.2 - Prob. 28ECh. 3.2 - Prob. 29ECh. 3.2 - Prob. 30ECh. 3.2 - Prob. 31ECh. 3.2 - Prob. 32ECh. 3.2 - Prob. 33ECh. 3.2 - Prob. 34ECh. 3.2 - Prob. 35ECh. 3.2 - Prob. 36ECh. 3.2 - Prob. 37ECh. 3.3 - Prob. 1ECh. 3.3 - Prob. 2ECh. 3.3 - Prob. 3ECh. 3.3 - Prob. 4ECh. 3.3 - Prob. 5ECh. 3.3 - Prob. 6ECh. 3.3 - Prob. 7ECh. 3.3 - Prob. 8ECh. 3.3 - Prob. 9ECh. 3.3 - Prob. 10ECh. 3.3 - Prob. 11ECh. 3.3 - Prob. 12ECh. 3.3 - Prob. 13ECh. 3.3 - Prob. 14ECh. 3.3 - Prob. 15ECh. 3.3 - Prob. 16ECh. 3.3 - Prob. 17ECh. 3.3 - Prob. 18ECh. 3.3 - Prob. 19ECh. 3.3 - Prob. 20ECh. 3.3 - Prob. 21ECh. 3.3 - Prob. 22ECh. 3.3 - Prob. 23ECh. 3.3 - Prob. 24ECh. 3.3 - Prob. 25ECh. 3.3 - Prob. 26ECh. 3.3 - Prob. 27ECh. 3.3 - Prob. 28ECh. 3.3 - Prob. 29ECh. 3.3 - Prob. 30ECh. 3.3 - Prob. 31ECh. 3.CR - Prob. 1CRCh. 3.CR - Prob. 2CRCh. 3.CR - Prob. 3CRCh. 3.CR - Prob. 4CRCh. 3.CR - Prob. 5CRCh. 3.CR - Prob. 6CRCh. 3.CR - Prob. 7CRCh. 3.CR - Prob. 8CRCh. 3.CR - Prob. 9CRCh. 3.CR - Prob. 10CRCh. 3.CR - Prob. 11CRCh. 3.CR - Prob. 12CRCh. 3.CR - Prob. 13CRCh. 3.CR - Prob. 14CRCh. 3.CR - Prob. 15CRCh. 3.PA - Prob. 1PCh. 3.PA - Prob. 2PCh. 3.PA - Prob. 3PCh. 3.PA - Prob. 4PCh. 3.PA - Prob. 5PCh. 3.PA - Prob. 6PCh. 3.PA - Prob. 7PCh. 3.PA - Prob. 8PCh. 3.PA - Prob. 9PCh. 3.PA - Prob. 10PCh. 3.PB - Prob. 1PCh. 3.PB - Prob. 2PCh. 3.PB - Prob. 3PCh. 3.PB - Prob. 4P
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