The scores of a midterm exam are normally distributed with a mean $\mu = 70$ and a standard deviation $\sigma = 15$. We want to find the percentage of students who score between 60 and 75. The given formula is: $P(a \le X \le b) = \int_a^b \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx$

1. Problem Description

The scores of a midterm exam are normally distributed with a mean $\mu = 70$ and a standard deviation $\sigma = 15$. We want to find the percentage of students who score between 60 and

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5. The given formula is:

$P(a \le X \le b) = \int_a^b \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx$

2. Solution Steps

First, we need to calculate the z-scores for 60 and 75 using the formula:

$z = \frac{x - \mu}{\sigma}$

For $x = 60$:

$z_1 = \frac{60 - 70}{15} = \frac{-10}{15} = -\frac{2}{3} \approx -0.67$

For $x = 75$:

$z_2 = \frac{75 - 70}{15} = \frac{5}{15} = \frac{1}{3} \approx 0.33$

Now we need to find the area under the standard normal curve between $z_1 = -0.67$ and $z_2 = 0.33$. We can look these values up in a standard normal distribution table (z-table).

$P(z < -0.67) = 0.2514$

$P(z < 0.33) = 0.6293$

To find the probability between these two z-scores, we subtract the smaller probability from the larger probability:

$P(-0.67 < z < 0.33) = P(z < 0.33) - P(z < -0.67) = 0.6293 - 0.2514 = 0.3779$

Converting this probability to a percentage:

$0.3779 \times 100\% = 37.79\%$

The closest answer to this is 0.378 or 37.8%.

3. Final Answer

C. 0.378