Question

the scores of an exam have a normal distribution. the mean of the scores is 48 and the standard deviation is 5. approximately what percent of the students taking the exam can be expected to score between 43 and 53? 68% 13% 34% 95%

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 48\) (mean) and \(\sigma=5\) (standard deviation).
For \(x = 43\): \(z_1=\frac{43 - 48}{5}=\frac{-5}{5}=- 1\)
For \(x = 53\): \(z_2=\frac{53 - 48}{5}=\frac{5}{5}=1\)

Step2: Use the empirical rule

The empirical rule (68 - 95 - 99.7 rule) for a normal distribution states that approximately \(P(\mu-\sigma<X<\mu + \sigma)=68\%\), where \(X\) is a normally distributed random variable. Here, \(\mu-\sigma=48 - 5=43\) and \(\mu+\sigma=48 + 5=53\)

Answer:

68%