Question

the exam scores of 10 students in a statistics class are listed below: 50 23 41 95 77 68 90 55 35 35. the standard deviation for the data is a. 23 b. 24.69 c. 23.42 d. 609.51

Explanation:

Step1: Recall standard - deviation formula

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$, where $x_{i}$ are the data - points, $\bar{x}$ is the sample mean, and $n$ is the sample size. First, find the mean $\bar{x}$.
The data set is $x=\{50,23,41,95,77,68,90,55,35,35\}$, and $n = 10$.
$\bar{x}=\frac{50 + 23+41+95+77+68+90+55+35+35}{10}=\frac{579}{10}=57.9$.

Step2: Calculate $(x_{i}-\bar{x})^{2}$ for each data - point

$(50 - 57.9)^{2}=(-7.9)^{2}=62.41$;
$(23 - 57.9)^{2}=(-34.9)^{2}=1218.01$;
$(41 - 57.9)^{2}=(-16.9)^{2}=285.61$;
$(95 - 57.9)^{2}=(37.1)^{2}=1376.41$;
$(77 - 57.9)^{2}=(19.1)^{2}=364.81$;
$(68 - 57.9)^{2}=(10.1)^{2}=102.01$;
$(90 - 57.9)^{2}=(32.1)^{2}=1030.41$;
$(55 - 57.9)^{2}=(-2.9)^{2}=8.41$;
$(35 - 57.9)^{2}=(-22.9)^{2}=524.41$;
$(35 - 57.9)^{2}=(-22.9)^{2}=524.41$.
The sum $\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=62.41+1218.01+285.61+1376.41+364.81+102.01+1030.41+8.41+524.41+524.41 = 5496.9$.

Step3: Calculate the sample standard deviation

$s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}=\sqrt{\frac{5496.9}{9}}\approx\sqrt{610.767}\approx24.69$.

Answer:

b. 24.69