Question

the scores on an exam had a mean of 65 and a standard deviation of 7.4. approximately what percentage of the students had a score over 75?

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 75\), \(\mu=65\), and \(\sigma = 7.4\).

$$z=\frac{75 - 65}{7.4}=\frac{10}{7.4}\approx1.35$$

Step2: Find the proportion using the standard normal distribution table

Looking up the z - score of \(z = 1.35\) in the standard normal distribution table (or using a calculator with a normal distribution function). The area to the left of \(z = 1.35\) is \(P(Z<1.35)\approx0.9115\).
The proportion of students with a score over \(75\) is \(P(X > 75)=1 - P(X\leq75)\). Since \(P(X\leq75)\) corresponds to the area to the left of \(z = 1.35\) in the standard normal distribution (after standardizing \(X\) to \(Z\)), \(P(X > 75)=1-0.9115 = 0.0885\).

Step3: Convert the proportion to a percentage

To convert the proportion to a percentage, we use the formula \(Percentage=Proportion\times100\%\).

$$Percentage=0.0885\times100\% = 8.85\%\approx9\%$$

Answer:

Approximately \(9\%\) of the students had a score over \(75\).