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.)

(a) 0.9699

(b) 0.8351

(c) = 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()

The z score probability distribution is given by;

          Z = ~ N(0,1)

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

    P(X < 88) = P( < ) = 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 65)

    P(X < 93) = P( < ) = P(Z < 2.50) = 0.99379

    P(X 65) = P( ) = P(Z -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( > ) = 0.25

     P(Z > ) = 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.;

            = 0.6745

              - 73 =

                = 73 + 5.396 = 78.396 ≈ 78

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

1. minimum

2. 6400

3. eight

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