Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 6.CR, Problem 15CR
To determine

a.

To find:

The probability of random variable less than given x-value.

Expert Solution
Check Mark

Answer to Problem 15CR

Solution:

The required probability is 0.8665.

Explanation of Solution

Given:

μ=22.0

σ=1.8

x=24.0

Description:

In statistics, a measurable function that is used to represent outcomes or occurrences of random phenomenon as real numbers is defined as a random variable.

The normal distribution is the most prevalent continuous probability distribution function and the real world hosts a milieu of the variables that are normally distributed.

In the normal distribution (also known as Gaussian or Gauss-Laplacian) distribution, the probability density function is given by:

f(x|μ,σ2)=12πσ2e(xμ)22σ2

Where, x is the value of the random variable while μ is the mean and the variance of the random variable which is normally distributed as σ2.

The normal Gaussian distribution with the following characteristics:

μ=0, and

σ=1

is called standard normal distribution.

The appropriate probability density function transforms to

f(x|0,1)=12πex22

A normal distribution can be converted to standard normal distribution by defining the z-score that is represented by using the following transformation:

z=xμσ

where x is the value of random variable and μ and σ are the mean and standard deviation of original normal distribution respectively.

The probabilities being calculated using the x- and z-values are related as below:

P(Xx)=P(Zz).

Calculation:

Given value of random variable (x), mean (μ) and standard deviation (σ), then z-value (z) is obtained using:

z=xμσ.

Thereafter, the relation P(Xx)=P(Zz) is used to find the required probability.

First, the z-value is computed:

z=(24.0)(22.0)(1.8)=2.01.8=1.11

Then, from the table:

P(Zz)=0.86650,

which in turn gives:

P(Xx)=0.8665.

Conclusion:

The required probability is 0.8665.

To determine

b.

To find:

Value of the mean.

Expert Solution
Check Mark

Answer to Problem 15CR

Solution:

The value of the mean is 116.11.

Explanation of Solution

Given:

σ=0.24

x=116.00

P(Xx)=0.3228

Description:

In statistics, a measurable function that is used to represent outcomes or occurrences of random phenomenon as real numbers is defined as a random variable.

The normal distribution is the most prevalent continuous probability distribution function and the real world hosts a milieu of the variables that are normally distributed.

In the normal distribution (also known as Gaussian or Gauss-Laplacian) distribution, the probability density function is given by:

f(x|μ,σ2)=12πσ2e(xμ)22σ2

Where, x is the value of the random variable while μ is the mean and the variance of the random variable which is normally distributed as σ2.

The normal Gaussian distribution with the following characteristics:

μ=0, and

σ=1

is called standard normal distribution.

The appropriate probability density function transforms to

f(x|0,1)=12πex22

A normal distribution can be converted to standard normal distribution by defining the z-score that is represented by using the following transformation:

z=xμσ

where x is the value of random variable and μ and σ are the mean and standard deviation of original normal distribution respectively.

Calculation:

Given value of probability P(Xx)=P(Zz), x-value (x) and standard deviation (σ), then mean (μ) is obtained using:

μ=xz×σ,

before which the z-value is looked up from the standard table.

The z-value as from the standard normal table is as:

z=0.46

Here, the mean has to be computed:

μ=(116.00)(0.46)×(0.24)=116.00(0.1104)=116.00+0.1104=116.1104

Rounding off to two decimals places gives:

μ=116.11.

Conclusion:

The value of the mean is 116.11.

To determine

c.

To find:

The standard deviation for given data.

Expert Solution
Check Mark

Answer to Problem 15CR

Solution:

The required standard deviation is 0.25.

Explanation of Solution

Given:

μ=0.46

x=0.72

P(Xx)=0.85

Description:

In statistics, a measurable function that is used to represent outcomes or occurrences of random phenomenon as real numbers is defined as a random variable.

The normal distribution is the most prevalent continuous probability distribution function and the real world hosts a milieu of the variables that are normally distributed.

In the normal distribution (also known as Gaussian or Gauss-Laplacian) distribution, the probability density function is given by:

f(x|μ,σ2)=12πσ2e(xμ)22σ2

Where, x is the value of the random variable while μ is the mean and the variance of the random variable which is normally distributed as σ2.

The normal Gaussian distribution with the following characteristics:

μ=0, and

σ=1

is called standard normal distribution.

The appropriate probability density function transforms to

f(x|0,1)=12πex22

A normal distribution can be converted to standard normal distribution by defining the z-score that is represented by using the following transformation:

z=xμσ

where x is the value of random variable and μ and σ are the mean and standard deviation of original normal distribution respectively.

Calculation:

Given the value of probability P(Xx)=P(Zz), value of random variable (x) and mean (μ), then standard deviation (σ) is obtained using:

σ=xμz,

before which the z-value is looked up from the standard table.

The z-value as from the standard normal table is:

z=1.04

Here, the standard deviation is to be calculated:

σ=(0.72)(0.46)(1.04)=0.261.04=0.25

This value rounded off to the 3rd decimal place gives:

σ=0.25

Conclusion:

The required standard deviation is 0.25.

To determine

d.

To find:

The x-value for given data.

Expert Solution
Check Mark

Answer to Problem 15CR

Solution:

The required x-value is 972.

Explanation of Solution

Given:

μ=1052

σ=66

P(Xx)=0.1131

Description:

In statistics, a measurable function that is used to represent outcomes or occurrences of random phenomenon as real numbers is defined as a random variable.

The normal distribution is the most prevalent continuous probability distribution function and the real world hosts a milieu of the variables that are normally distributed.

In the normal distribution (also known as Gaussian or Gauss-Laplacian) distribution, the probability density function is given by:

f(x|μ,σ2)=12πσ2e(xμ)22σ2

Where, x is the value of the random variable while μ is the mean and the variance of the random variable which is normally distributed as σ2.

The normal Gaussian distribution with the following characteristics:

μ=0, and

σ=1

is called standard normal distribution.

The appropriate probability density function transforms to

f(x|0,1)=12πex22

A normal distribution can be converted to standard normal distribution by defining the z-score that is represented by using the following transformation:

z=xμσ

where x is the value of random variable and μ and σ are the mean and standard deviation of original normal distribution respectively.

Calculation:

Given the value of probability P(Xx)=P(Zz), standard deviation (σ) and mean (μ), then value of random variable (x) is obtained using:

x=μ+σ×z

before which the z-value is looked up from the standard table.

The z-value from the table is:

z=1.21.

Then, the x-value, that is the value of the random variable is to be obtained by:

x=(1052)+(66)×(1.21)=1052+(79.86)=105279.86=972.14

This number rounded off to the nearest whole number gives:

x=972.

Conclusion:

The required x-value is 972.

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Chapter 6 Solutions

Beginning Statistics, 2nd Edition

Ch. 6.1 - Prob. 11ECh. 6.1 - Prob. 12ECh. 6.1 - Prob. 13ECh. 6.1 - Prob. 14ECh. 6.1 - Prob. 15ECh. 6.1 - Prob. 16ECh. 6.1 - Prob. 17ECh. 6.1 - Prob. 18ECh. 6.1 - Prob. 19ECh. 6.1 - Prob. 20ECh. 6.1 - Prob. 21ECh. 6.1 - Prob. 22ECh. 6.1 - Prob. 23ECh. 6.1 - Prob. 24ECh. 6.1 - Prob. 25ECh. 6.2 - Prob. 1ECh. 6.2 - Prob. 2ECh. 6.2 - Prob. 3ECh. 6.2 - Prob. 4ECh. 6.2 - Prob. 5ECh. 6.2 - Prob. 6ECh. 6.2 - Prob. 7ECh. 6.2 - Prob. 8ECh. 6.2 - Prob. 9ECh. 6.2 - Prob. 10ECh. 6.2 - Prob. 11ECh. 6.2 - Prob. 12ECh. 6.2 - Prob. 13ECh. 6.2 - Prob. 14ECh. 6.2 - Prob. 15ECh. 6.2 - Prob. 16ECh. 6.2 - Prob. 17ECh. 6.2 - Prob. 18ECh. 6.2 - Prob. 19ECh. 6.2 - Prob. 20ECh. 6.2 - Prob. 21ECh. 6.2 - Prob. 22ECh. 6.2 - Prob. 23ECh. 6.2 - Prob. 24ECh. 6.2 - Prob. 25ECh. 6.2 - Prob. 26ECh. 6.2 - Prob. 27ECh. 6.2 - Prob. 28ECh. 6.2 - Prob. 29ECh. 6.2 - Prob. 30ECh. 6.2 - Prob. 31ECh. 6.2 - Prob. 32ECh. 6.2 - Prob. 33ECh. 6.2 - Prob. 34ECh. 6.2 - Prob. 35ECh. 6.2 - Prob. 36ECh. 6.2 - Prob. 37ECh. 6.2 - Prob. 38ECh. 6.2 - Prob. 39ECh. 6.2 - Prob. 40ECh. 6.2 - Prob. 41ECh. 6.2 - Prob. 42ECh. 6.2 - Prob. 43ECh. 6.2 - Prob. 44ECh. 6.2 - Prob. 45ECh. 6.2 - Prob. 46ECh. 6.2 - Prob. 47ECh. 6.2 - Prob. 48ECh. 6.2 - Prob. 49ECh. 6.3 - Prob. 1ECh. 6.3 - Prob. 2ECh. 6.3 - Prob. 3ECh. 6.3 - Prob. 4ECh. 6.3 - Prob. 5ECh. 6.3 - Prob. 6ECh. 6.3 - Prob. 7ECh. 6.3 - Prob. 8ECh. 6.3 - Prob. 9ECh. 6.3 - Prob. 10ECh. 6.3 - Prob. 11ECh. 6.4 - Prob. 1ECh. 6.4 - Prob. 2ECh. 6.4 - Prob. 3ECh. 6.4 - Prob. 4ECh. 6.4 - Prob. 5ECh. 6.4 - Prob. 6ECh. 6.4 - Prob. 7ECh. 6.4 - Prob. 8ECh. 6.4 - Prob. 9ECh. 6.4 - Prob. 10ECh. 6.4 - Prob. 11ECh. 6.4 - Prob. 12ECh. 6.4 - Prob. 13ECh. 6.4 - Prob. 14ECh. 6.4 - Prob. 15ECh. 6.4 - Prob. 16ECh. 6.4 - Prob. 17ECh. 6.4 - Prob. 18ECh. 6.4 - Prob. 19ECh. 6.4 - Prob. 20ECh. 6.4 - Prob. 21ECh. 6.4 - Prob. 22ECh. 6.4 - Prob. 23ECh. 6.4 - Prob. 24ECh. 6.4 - Prob. 25ECh. 6.4 - Prob. 26ECh. 6.4 - Prob. 27ECh. 6.4 - Prob. 28ECh. 6.4 - Prob. 29ECh. 6.5 - Prob. 1ECh. 6.5 - Prob. 2ECh. 6.5 - Prob. 3ECh. 6.5 - Prob. 4ECh. 6.5 - Prob. 5ECh. 6.5 - Prob. 6ECh. 6.5 - Prob. 7ECh. 6.5 - Prob. 8ECh. 6.5 - Prob. 9ECh. 6.5 - Prob. 10ECh. 6.5 - Prob. 11ECh. 6.5 - Prob. 12ECh. 6.5 - Prob. 13ECh. 6.5 - Prob. 14ECh. 6.5 - Prob. 15ECh. 6.5 - Prob. 16ECh. 6.5 - Prob. 17ECh. 6.5 - Prob. 18ECh. 6.5 - Prob. 19ECh. 6.5 - Prob. 20ECh. 6.5 - Prob. 21ECh. 6.5 - Prob. 22ECh. 6.5 - Prob. 23ECh. 6.5 - Prob. 24ECh. 6.5 - Prob. 25ECh. 6.5 - Prob. 26ECh. 6.5 - Prob. 27ECh. 6.5 - Prob. 28ECh. 6.5 - Prob. 29ECh. 6.5 - Prob. 30ECh. 6.CR - Prob. 1CRCh. 6.CR - Prob. 2CRCh. 6.CR - Prob. 3CRCh. 6.CR - Prob. 4CRCh. 6.CR - Prob. 5CRCh. 6.CR - Prob. 6CRCh. 6.CR - Prob. 7CRCh. 6.CR - Prob. 8CRCh. 6.CR - Prob. 9CRCh. 6.CR - Prob. 10CRCh. 6.CR - Prob. 11CRCh. 6.CR - Prob. 12CRCh. 6.CR - Prob. 13CRCh. 6.CR - Prob. 14CRCh. 6.CR - Prob. 15CRCh. 6.CR - Prob. 16CRCh. 6.CR - Prob. 17CRCh. 6.CR - Prob. 18CRCh. 6.CR - Prob. 19CRCh. 6.CR - Prob. 20CRCh. 6.CR - Prob. 21CRCh. 6.CR - Prob. 22CRCh. 6.P - Prob. 1PCh. 6.P - Prob. 2PCh. 6.P - Prob. 3PCh. 6.P - Prob. 4PCh. 6.P - Prob. 5PCh. 6.P - Prob. 6PCh. 6.P - Prob. 7PCh. 6.P - Prob. 8P
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