A set of computer science exam scores are normally distributed with a mean of 71.33 point, and a standard deviation of 3 points. What proportion of the exam scores are between 68 and 77.99 points?

0.86 of the exam scores are between 68 and 77.99 points

Step-by-step explanation:

Since the set of computer science exam scores are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = computer science exam scores .

µ = mean score

σ = standard deviation

From the information given,

µ = 71.33 points

σ = 3 points

We want to find the proportion of the exam scores are between 68 and 77.99 points. It is expressed as

P(68 ≤ x ≤ 77.99)

For x = 68,

z = (68 - 71.33)/3 = - 1.11

Looking at the normal distribution table, the probability corresponding to the z score is 0.13

For x = 68,

z = (77.99 - 71.33)/3 = 2.22

Looking at the normal distribution table, the probability corresponding to the z score is 0.99

P(68 ≤ x ≤ 77.99) = 0.99 - 0.13 = 0.86

required rate of return = 0.166 = 16.6%

step-by-step explanation:

under the capital asset pricing model (capm), the required or expected rate of return for a particular security or asset is given using the formula:

required rate of return = risk-free rate + beta*(market risk premium)

we are given that;

risk-free rate = 4% = 0.04

beta = 1.4

market risk premium = 9% = 0.09

we simply plug these values into the above formula;

required rate of return = 0.04 + 1.4*(0.09)

required rate of return = 0.04 + 0.126

required rate of return = 0.166 = 16.6%

the answer is d

hope this