Question

A set of final examination grades in an introductory statistics course is normally​ distributed, with a mean of 73 and a standard deviation of 8. Complete parts​ (a) through​ (c). a. What is the probability that a student scored below 88 on this​ exam? The probability that a student scored below 88 is 0.9699. ​(Round to four decimal places as​ needed.) b. What is the probability that a student scored between 65 and 93​? The probability that a student scored between 65 and 93 is 0.8351. ​(Round to four decimal places as​ needed.) c. The probability is 25​% that a student taking the test scores higher than what​ grade? The probability is 25​% that a student taking the test scores higher than 78. ​(Round to the nearest integer as​ needed.)

Answer 1

Answer:

(a) 0.9699

(b) 0.8351

(c) $x$ = 78

Step-by-step explanation:

We are given that a set of final examination grades in an introductory statistics course is normally​ distributed, with a mean of 73 and a standard deviation of 8.

Let X = marks of students in an exam

So, X ~ N($\mu = 73, \sigma^{2}= 8^{2}$)

The z score probability distribution is given by;

          Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that a student scored below 88 on this​ exam = P(X < 88)

    P(X < 88) = P( $\frac{X-\mu}{\sigma}$ < $\frac{88-73}{8}$) = P(Z < 1.88) = 0.9699

(b) Probability that a student scored between 65 and 93 = P(65 < X < 93)

    P(65 < X < 93) = P(X < 93) - P(X $\leq$ 65)

    P(X < 93) = P( $\frac{X-\mu}{\sigma}$ < $\frac{93-73}{8}$) = P(Z < 2.50) = 0.99379

    P(X $\leq$ 65) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{65-73}{8}$) = P(Z $\leq$ -1) = 1 - P(Z < 1) = 1 - 0.84134 = 0.15866

Therefore, P(65 < X < 93) = 0.99379 - 0.15866 = 0.8351

(c) We have to find that probability is 25​% that a student taking the test scores higher than what​ grade, which means ;

    P(X > x) = 0.25

    P( $\frac{X-\mu}{\sigma}$ > $\frac{x-73}{8}$ ) = 0.25

     P(Z > $\frac{x-73}{8}$ ) = 0.25

Now in the z table the critical value of X which have an area greater than 0.25 is 0.6745, i.e.;

            $\frac{x-73}{8}$ = 0.6745

             $x$ - 73 = $0.6745 \times 8$

                $x$ = 73 + 5.396 = 78.396 ≈ 78

So, the probability is 25​% that a student taking the test scores higher than 78.