A random sample of math majors taking an introductory statistics course were surveyed after completing the final exam. They were asked, "How many times did you review your final exam before handing it in to the professor?" The results are displayed in a probability density function for the random variable X, the number of times students reviewed their exam before handing it in. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 1 1/5 2 2/5 7 2/5

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dfbustos dfbustos · Answer 1 · 2020-06-13T03:05:10+02:00

Answer:

[tex] E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8[/tex]

Now we can find the second moment with this formula:

[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]

And replacing we got:

[tex] E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4[/tex]

The variance would be given by:

[tex] Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96[/tex]

And the deviation would be:

[tex] Sd(X) =\sqrt{6.96}= 2.638[/tex]

Step-by-step explanation:

For this case we have the following distribution given:

X 1 2 7

P(X) 1/5 2/5 2/5

We need to begin finding the mean with this formula:

[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]

And replacing we got:

[tex] E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8[/tex]

Now we can find the second moment with this formula:

[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]

And replacing we got:

[tex] E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4[/tex]

The variance would be given by:

[tex] Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96[/tex]

And the deviation would be:

[tex] Sd(X) =\sqrt{6.96}= 2.638[/tex]