Question

the scores of a recent test taken by 450 students had an approximately normal distribution with a mean of 74 and a standard deviation of 3.6. determine the number of students who scored between 71 and 83. round your answer to the nearest whole number.

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the data - point.
For $x = 71$, $z_1=\frac{71 - 74}{3.6}=\frac{- 3}{3.6}\approx - 0.83$.
For $x = 83$, $z_2=\frac{83 - 74}{3.6}=\frac{9}{3.6}=2.5$.

Step2: Use the standard normal distribution table

The standard normal distribution table gives the cumulative probability $P(Z < z)$.
From the table, $P(Z < - 0.83)=0.2033$ and $P(Z < 2.5)=0.9938$.

Step3: Find the probability between the two z - scores

The probability $P(-0.83$P(-0.83

Step4: Calculate the number of students

The total number of students is $n = 450$.
The number of students with scores between 71 and 83 is $n\times P(-0.83

Answer:

356