Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
Question
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Chapter 5, Problem 4RE

a.

To determine

Construct the probability distribution for the random variable X.

a.

Expert Solution
Check Mark

Answer to Problem 4RE

The probability distribution for the random variable X is,

xP(x)
10.5632
20.2500
30.1147
40.0473
50.0171
60.0053
70.0018
80.0006

Explanation of Solution

Calculation:

The given table represents the frequency distribution of the number of AP tests taken by students who took one or more AP tests. The random variable X denotes the number of exams taken by a student who took one or more AP tests.

Here, the total number of people is 1,692. The corresponding probabilities of the random variable X are obtained by dividing the frequency of each hour (f) by total frequency (N).

That is, P(x)=fN

For x=1:

P(1)=9531,692=0.5632

Similarly the remaining probabilities are obtained as follows:

xfNP(x)
19531,6920.5632
24231,6920.2500
31941,6920.1147
4801,6920.0473
5291,6920.0171
691,6920.0053
731,6920.0018
811,6920.0006
Total 1.0000

Thus, the discrete probability distribution for the random variable X is obtained.

b.

To determine

Find the probability that a student took exactly one exam.

b.

Expert Solution
Check Mark

Answer to Problem 4RE

The probability that a student took exactly one exam is 0.5632.

Explanation of Solution

Calculation:

The table represents the probability distribution of the random variable X, the exams taken by a student who took one or more AP tests.

The probability that a student took exactly one exam is the probability at the point x=1. That is,

P(x=1)=0.5632

Thus, the probability that a student took exactly one exam is 0.5632.

c.

To determine

Find the mean.

c.

Expert Solution
Check Mark

Answer to Problem 4RE

The mean value is 1.7312.

Explanation of Solution

Calculation:

The formula for the mean of a discrete random variable is,

E(X)=μX=xP(x)

The mean of the random variable is obtained as given below:

xP(x)xP(x)
10.56320.5632
20.25000.5000
30.11470.3441
40.04730.1892
50.01710.0855
60.00530.0318
70.00180.0126
80.00060.0048
Total1.00001.7312

Thus, the mean value is 1.7312.

d.

To determine

Find the standard deviation.

d.

Expert Solution
Check Mark

Answer to Problem 4RE

The standard deviation is 1.049.

Explanation of Solution

Calculation:

The standard deviation of the random variable X is obtained by taking the square root of variance.

The formula for the variance of the discrete random variable X is,

σX2=[(xμX)2P(x)]

Where μX=[xP(x)] represents the mean of the random variable X.

The variance of the random variable X is obtained using the following table:

xP(x)(xμX)(xμX)2(xμX)2P(x)
10.5632–0.73120.5350.3011
20.25000.26880.0720.0181
30.11471.26881.6100.1847
40.04732.26885.1470.2435
50.01713.268810.6850.1827
60.00534.268818.2230.0966
70.00185.268827.7600.0500
80.00066.268839.2980.0236
Total1.000022.1504103.3301.1001

Therefore,

σ2=1.1001

Thus, the variance is 1.1001.

The standard deviation is,

σ=1.1001=1.049

That is, the standard deviation is 1.049.

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Chapter 5 Solutions

Essential Statistics

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