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23. Scores on a test are normally distributed with a mean of 63.2 and a standard deviation of 11.7. Find P81, which separates

Scores on a test are normally distributed with a mean of 63.2 and a standard deviation of 11.7. Find P81, which separates the bottom 81% from the top 19%. Round to two decimal places. 

A. 0.88 

B. 73.47 

C. 66.60 

D. 0.29

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Answer #2
Concepts and reason

Normal distribution:

Normal distribution is a continuous distribution of data that has the bell-shaped curve. The normally distributed random variable x has meanand variance (σ2)\left( {{\sigma ^2}} \right) .

Also, the standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean and standard deviation.

Standardized z-score:

The standardized z-score represents the number of standard deviations the data point is away from the mean.

• If the z-score takes positive value when it is above the mean (0).

• If the z-score takes negative value when it is below the mean (0).

Unbiased estimator:

An unbiased estimator is a statistic which estimates the parameter of interest with no bias. The expectation of the unbiased estimator results the parameter of interest.

Fundamentals

The test statistic with population standard deviation and population mean is,

z=xμσz = \frac{{x - \mu }}{\sigma }

Here,

x:Sampleunitμ:Populationmeanσ:Populationstandarddeviation\begin{array}{c}\\x:{\rm{Sample unit}}\\\\\mu :{\rm{Population mean}}\\\\\sigma :{\rm{Population standard deviation}}\\\end{array}

The sample mean is unbiased estimator of population mean.

The sample standard deviation is unbiased estimator of population standard deviation.

The test statistic with sample standard deviation and sample mean is,

z=xxˉsz = \frac{{x - \bar x}}{s}

Here,

x:Sampleunitxˉ:Samplemeans:Samplestandarddeviation\begin{array}{c}\\x:{\rm{Sample unit}}\\\\\bar x:{\rm{Sample mean}}\\\\s:{\rm{Sample standard deviation}}\\\end{array}

The critical value is obtained by ‘Standard normal table’.

Procedure for finding the Z-value is listed below:

1.From the table of standard normal distribution, locate the probability value.

2.Move left until the first column is reached.

3.Move upward until the top row is reached.

4.Locate the probability value, by the intersection of the row and column values gives the area to the left of z.

The z-score that corresponds to bottom 81% from the top 19% is obtained below.

From the given information, P(Zz)=0.81P\left( {Z \ge z} \right) = 0.81 .

Procedure for finding the z-value is listed below:

1.From the table of standard normal distribution, locate the probability value of 0.81.

2.Move left until the first column is reached. Note the value as 0.8

3.Move upward until the top row is reached. Note the value as 0.08.

4.The intersection of the row and column values gives the area to the two tail of z.

That is, P(Z0.88)=0.81P\left( {Z \ge 0.88} \right) = 0.81 .

From standard normal table, the required Z0.81{Z_{0.81}} is 0.88.

The P81{P_{81}} which that separates the bottom 81% from the top 19% is obtained below:

From the given information, xˉ=63.2\bar x = 63.2 , s=11.7s = 11.7 and Z0.81=0.88{Z_{0.81}} = 0.88

The required value is,

z=xxˉsx=zsxˉx=0.88×11.7+63.2=10.296+63.20\begin{array}{c}\\z = \frac{{x - \bar x}}{s}\\\\x = zs - \bar x\\\\x = 0.88 \times 11.7 + 63.2\\\\ = 10.296 + 63.20\\\end{array}

=73.49673.5\begin{array}{l}\\ = 73.496\\\\ \simeq 73.5\\\end{array}

Ans:

The P81{P_{81}} which that separates the bottom 81% from the top 19% is 73.5.

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Answer #1

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 63.2

S.d = 11.7

We need to find

P81

From z table, P(z<0.88) = 81%

So, 0.88 = (x - 63.2)/11.7

X = 73.496

Closest option is 73.47

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