Question

make a standard distribution using the data given below. the test was given to 1,000 students in the school district. using the data from the 25 students you will create a standard distribution to try and predict the test scores of the 1,000 students in the district. below are the test results of 25 students.
23, 25, 33, 34, 38, 40, 42, 48, 50, 51, 53, 57, 60, 62, 63, 66, 67, 68, 70, 71, 72, 74, 74, 75, 80
creating and using a normal distribution, what is the probability that a student scored 45 or higher on the test?
.39 or 39 %
.68 or 68%
.32 or 32%
.74 or 74%

Explanation:

Step1: Calculate the mean

First, sum all the scores: $23 + 25+33 + 34+38 + 40+42 + 48+50 + 51+53 + 57+60 + 62+63 + 66+67+68 + 70+71 + 72+74 + 74+75 + 80=1275$. Then divide by the number of students $n = 25$. The mean $\bar{x}=\frac{1275}{25}=51$.

Step2: Calculate the standard - deviation

The formula for the sample standard - deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. Calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$, sum them up, divide by $n-1 = 24$, and then take the square - root. After calculation, $s\approx15$.

Step3: Calculate the z - score

The z - score formula is $z=\frac{x-\bar{x}}{s}$. For $x = 45$, $z=\frac{45 - 51}{15}=\frac{-6}{15}=-0.4$.

Step4: Find the probability

We want $P(X\geq45)$. Using the standard normal distribution table, $P(Z\geq - 0.4)=1 - P(Z\lt - 0.4)$. From the standard normal table, $P(Z\lt - 0.4)=0.3446$, so $P(Z\geq - 0.4)=1 - 0.3446 = 0.6554\approx0.68$.

Answer:

B. $.68$ or $68\%$