Question

An English professor assigns letter grades on a test according to the following scheme. A: Top 5% of scores B: Scores below the top 5% and above the bottom 64% C: Scores below the top 36% and above the bottom 23% D: Scores below the top 77% and above the bottom 9% F: Bottom 9% of scores Scores on the test are normally distributed with a mean of 78.8 and a standard deviation of 7.1. Find the numerical limits for a D grade. Round your answers to the nearest whole number, if necessary.

Answer 1

Answer:

Grade D score:

$69 \leq x \leq 74$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 78.8

Standard Deviation, σ = 7.1

We are given that the distribution of scores on test is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

D: Scores below the top 77% and above the bottom 9%

We have to find the value of x such that the probability is 0.77

$P( X > x) = P( z > \displaystyle\frac{x - 78.8}{7.1})=0.77$  

$= 1 -P( z \leq \displaystyle\frac{x - 78.8}{7.1})=0.77$  

$=P( z \leq \displaystyle\frac{x - 78.8}{7.1})=0.23$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 78.8}{7.1} =-0.739\\\\x = 73.55$  

We have to find the value of x such that the probability is 0.09

$P(X < 0.09) = \\\\P( X < x) = P( z < \displaystyle\frac{x - 78.8}{7.1})=0.09$

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 78.8}{7.1} = -1.341\\\\x = 69.27$  

Thus, the numerical value of score to achieve grade D is

$69 \leq x \leq 74$