6th Edition

ISBN: 9781319113339

Author: Starnes

Publisher: MAC HIGHER

Concept explainers

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

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Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 1.3, Problem 101E

(a)

To determine

To interpret the standard deviation.

(a)

Expert Solution

Answer to Problem 101E

Standard deviation = 1.67

Explanation of Solution

Given:

Mean = 3.2

Standard deviation = 1.9

PRACTICE OF STATISTICS F/AP EXAM, Chapter 1.3, Problem 101E

The standard deviation defines the deviation or variation of the data values from the mean.. The standard deviation shows the average distance between the each data value and the mean, which means that 1.9 is the average distance between the data value and the mean.

(b)

To determine

To explain the effect of change in the observation from 22 to 35.

(b)

Expert Solution

Answer to Problem 101E

Both mean and standard deviation will decrease.

Explanation of Solution

Since the new value 4 is smaller than the previous value 7.5, so the mean will decrease and since the previous value 7.5 is far from the mean comparatively to the new value 4, which means the value of deviation sum of square will decrease now and that ultimately will decrease the value of standard deviation.

Chapter 1.3, Problem 100E

Chapter 1.3, Problem 102E

Chapter 1 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E

Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.2 - Prob. 45E Ch. 1.2 - Prob. 46E Ch. 1.2 - Prob. 47E Ch. 1.2 - Prob. 48E Ch. 1.2 - Prob. 49E Ch. 1.2 - Prob. 50E Ch. 1.2 - Prob. 51E Ch. 1.2 - Prob. 52E Ch. 1.2 - Prob. 53E Ch. 1.2 - Prob. 54E Ch. 1.2 - Prob. 55E Ch. 1.2 - Prob. 56E Ch. 1.2 - Prob. 57E Ch. 1.2 - Prob. 58E Ch. 1.2 - Prob. 59E Ch. 1.2 - Prob. 60E Ch. 1.2 - Prob. 61E Ch. 1.2 - Prob. 62E Ch. 1.2 - Prob. 63E Ch. 1.2 - Prob. 64E Ch. 1.2 - Prob. 65E Ch. 1.2 - Prob. 66E Ch. 1.2 - Prob. 67E Ch. 1.2 - Prob. 68E Ch. 1.2 - Prob. 69E Ch. 1.2 - Prob. 70E Ch. 1.2 - Prob. 71E Ch. 1.2 - Prob. 72E Ch. 1.2 - Prob. 73E Ch. 1.2 - Prob. 74E Ch. 1.2 - Prob. 75E Ch. 1.2 - Prob. 76E Ch. 1.2 - Prob. 77E Ch. 1.2 - Prob. 78E Ch. 1.2 - Prob. 79E Ch. 1.2 - Prob. 80E Ch. 1.2 - Prob. 81E Ch. 1.2 - Prob. 82E Ch. 1.2 - Prob. 83E Ch. 1.2 - Prob. 84E Ch. 1.2 - Prob. 85E Ch. 1.2 - Prob. 86E Ch. 1.3 - Prob. 87E Ch. 1.3 - Prob. 88E Ch. 1.3 - Prob. 89E Ch. 1.3 - Prob. 90E Ch. 1.3 - Prob. 91E Ch. 1.3 - Prob. 92E Ch. 1.3 - Prob. 93E Ch. 1.3 - Prob. 94E Ch. 1.3 - Prob. 95E Ch. 1.3 - Prob. 96E Ch. 1.3 - Prob. 97E Ch. 1.3 - Prob. 98E Ch. 1.3 - Prob. 99E Ch. 1.3 - Prob. 100E Ch. 1.3 - Prob. 101E Ch. 1.3 - Prob. 102E Ch. 1.3 - Prob. 103E Ch. 1.3 - Prob. 104E Ch. 1.3 - Prob. 105E Ch. 1.3 - Prob. 106E Ch. 1.3 - Prob. 107E Ch. 1.3 - Prob. 108E Ch. 1.3 - Prob. 109E Ch. 1.3 - Prob. 110E Ch. 1.3 - Prob. 111E Ch. 1.3 - Prob. 112E Ch. 1.3 - Prob. 113E Ch. 1.3 - Prob. 114E Ch. 1.3 - Prob. 115E Ch. 1.3 - Prob. 116E Ch. 1.3 - Prob. 117E Ch. 1.3 - Prob. 118E Ch. 1.3 - Prob. 119E Ch. 1.3 - Prob. 120E Ch. 1.3 - Prob. 121E Ch. 1.3 - Prob. 122E Ch. 1.3 - Prob. 123E Ch. 1.3 - Prob. 124E Ch. 1.3 - Prob. 125E Ch. 1.3 - Prob. 126E Ch. 1.3 - Prob. 127E Ch. 1.3 - Prob. 128E Ch. 1 - Prob. 1E Ch. 1 - Prob. 2E Ch. 1 - Prob. 3E Ch. 1 - Prob. 4E Ch. 1 - Prob. 5E Ch. 1 - Prob. 6E Ch. 1 - Prob. 7E Ch. 1 - Prob. 8E Ch. 1 - Prob. 9E Ch. 1 - Prob. 10E Ch. 1 - Prob. R1.1RE Ch. 1 - Prob. R1.2RE Ch. 1 - Prob. R1.3RE Ch. 1 - Prob. R1.4RE Ch. 1 - Prob. R1.5RE Ch. 1 - Prob. R1.6RE Ch. 1 - Prob. R1.7RE Ch. 1 - Prob. R1.8RE Ch. 1 - Prob. R1.9RE Ch. 1 - Prob. R1.10RE Ch. 1 - Prob. T1.1SPT Ch. 1 - Prob. T1.2SPT Ch. 1 - Prob. T1.3SPT Ch. 1 - Prob. T1.4SPT Ch. 1 - Prob. T1.5SPT Ch. 1 - Prob. T1.6SPT Ch. 1 - Prob. T1.7SPT Ch. 1 - Prob. T1.8SPT Ch. 1 - Prob. T1.9SPT Ch. 1 - Prob. T1.10SPT Ch. 1 - Prob. T1.11SPT Ch. 1 - Prob. T1.12SPT Ch. 1 - Prob. T1.13SPT Ch. 1 - Prob. T1.14SPT

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