Question

Repeat part $\left(a\right)$of Problem$8.2$when it is known that the variance of a student’s test score is equal

to$25$.

Short answer

Therefore,

$P\{X\ge 85\}\le \frac{1}{5}=.2$.

Step-by-step solution

Given Information.

the variance of a student’s test score is equal to$25$.

Explanation.

Let $X$represents the test score of a student taking her final examination. Assume that $X$is a random variable with mean $\mu =75$and variance${\sigma }^2$$=25$.

Then, using Corollarylocalid="1649833457509" $5.1$. (textbook)$\underset{¯}{a=10}$, we get:

localid="1650364326770" $P\{X\ge 85\}=P\{X\ge 75+10\}\le \frac{{\sigma }^2}{{\sigma }^2+{10}^2}=\frac{25}{25+100}=\frac{1}{5}=0.2$