Question

Math 132 final exam grade is normally distributed with mean 68 and standard deviation 23. Final exam score above 92 corresponds to an A Approximately what percent of the class gotan A?

Answer 1

The percentage of the class that scored an A on the Math 132 final exam is calculated using the z-score for a score of 92. By finding the corresponding percentile for the z-score and subtracting from 100%, approximately 15% of the class got an A.

Step-by-step explanation:

To determine what percent of the class scored an A (a score above 92) on the Math 132 final exam, we need to calculate the z-score and then find the corresponding percentile. The exam scores are normally distributed with a mean (μ) of 68 and a standard deviation (σ) of 23. The z-score is calculated using the formula: z = (X - μ) / σ, where X is the score in question.

To calculate the z-score for an A (score of 92): z = (92 - 68) / 23 = 1.043. Next, we can consult the standard normal distribution table, or use technology, to find the percentile that corresponds to a z-score of 1.043. This percentile tells us the percentage of students who scored below 92. To find the percentage who scored above 92, we subtract this value from 100%.

Assuming a z-score of 1.043 corresponds approximately to a percentile of 85, this would mean that about 15% of the class received an A.

Answer 2

Since the function is a normal distribution, we can use the z-score formula shown below

`Z=(x-\mu)/(\sigma)`

In our case,

`\mu=68,\sigma=23`

Then, set x=92 and solve for Z

`\begin{gathered} x=92 \\ \Rightarrow Z=(92-68)/(23)=(24)/(23) \\ \Rightarrow Z=(24)/(23)=1.04347\ldots \end{gathered}`

Using a z-score table, the cumulative probability of Z=24/23 is

`\Rightarrow P(X\le92)=0.8508`

Finally,

`\begin{gathered} \Rightarrow P(X>92)=1-P(X\le92)=1-0.8508=0.1492 \\ \Rightarrow P(X>92)=0.1492 \end{gathered}`

Thus, the answer is 0.1492, which is equivalent to 14.92%; rounded to the nearest percentage, the answer is 15%