Question

1. (4 points) suppose the scores of students on a test are (approximately) normally distributed with a mean score of 70 points, and a standard deviation of 10 points. it is decided by the committee to give the grades as to 10 percent of the students, and grades bs to 23 percent of the students. find what scores should be assigned for grades as and bs?

Explanation:

Step1: Find the z - score for the top 10%

Using the standard normal distribution table (or z - table), the z - score \(z\) such that \(P(Z>z)=0.10\). Since \(P(Z > z)=0.10\), then \(P(Z\leq z)=1 - 0.10=0.90\). Looking up in the standard normal table, the \(z\) - score corresponding to a cumulative probability of \(0.90\) is approximately \(z = 1.28\).

Step2: Use the z - score formula \(z=\frac{x-\mu}{\sigma}\)

We know that \(\mu = 70\) (mean), \(\sigma=10\) (standard deviation), and \(z = 1.28\). Rearranging the formula \(z=\frac{x-\mu}{\sigma}\) for \(x\) gives \(x=\mu+z\sigma\).
Substitute the values: \(x = 70+1.28\times10\).

$$x=70 + 12.8=82.8$$

Step3: Find the z - score for the top 23%

The \(z\) - score \(z\) such that \(P(Z>z)=0.23\). Then \(P(Z\leq z)=1 - 0.23 = 0.77\). Looking up in the standard normal table, the \(z\) - score corresponding to a cumulative probability of \(0.77\) is approximately \(z = 0.74\).

Step4: Use the z - score formula \(z=\frac{x-\mu}{\sigma}\) again

Substitute \(\mu = 70\), \(\sigma = 10\), and \(z = 0.74\) into \(x=\mu+z\sigma\).

$$x=70+0.74\times10=70 + 7.4 = 77.4$$

Answer:

Scores of \(82.8\) and above should be assigned grade A's for the top 10% of students, and scores of \(77.4\) and above should be considered for the top 23% of students (assuming the committee uses the calculated cut - offs based on the normal distribution with \(\mu = 70\) and \(\sigma=10\)).