Question

the scores of a recent test taken by 1200 students had an approximately normal distribution with a mean of 225 and a standard deviation of 18. determine the number of students who scored less than 171.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 171$, $\mu=225$, and $\sigma = 18$.
Substitute the values into the formula: $z=\frac{171 - 225}{18}=\frac{- 54}{18}=- 3$.

Step2: Find the probability for $z=-3$

For a standard normal distribution, the cumulative probability $P(Z < - 3)$ can be found using the standard normal table or a calculator. The probability that $Z < - 3$ is approximately $0.0013$.

Step3: Calculate the number of students

The total number of students $N = 1200$. The number of students who scored less than 171 is $n=N\times P(Z < - 3)$.
Substitute $N = 1200$ and $P(Z < - 3)=0.0013$ into the formula: $n = 1200\times0.0013 = 1.56$. Wait, but the calculator shows $\frac{1200 - 225}{18}=\frac{975}{18}\approx54.17$? Wait, maybe there is a mis - reading of the problem. Wait, if we consider the formula in the calculator: $\frac{1200 - 225}{18}=\frac{975}{18}\approx54.17$. Maybe the problem was mis - stated, and we are just calculating $\frac{1200 - 225}{18}$. Let's recalculate that: $1200-225 = 975$, then $\frac{975}{18}=\frac{325}{6}\approx54.17$.

Answer:

Approximately 54.17 (If we follow the calculation in the calculator $\frac{1200 - 225}{18}$) or 2 (if we follow the normal distribution calculation, but there is a discrepancy. If we follow the calculator's operation, the answer is approximately 54.17)