Question

students in class a and class b were given the same quiz. class a had a mean score of 8 points with a standard deviation of 0.8 points. class b had a mean score of 8.3 points with a standard deviation of 1.1 points. which class scored better on average? which class had more consistent scores? question 6 a sample was done, collecting the data below. calculate the standard deviation, to one decimal place. x 12 14 6 5 19

Explanation:

Step 1: Calculate the mean ($\bar{x}$) of the data set.

The data set is \( x = [12, 14, 6, 5, 19] \). The number of data points \( n = 5 \).
The formula for the mean is \( \bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n} \).
\( \sum_{i=1}^{5}x_{i}=12 + 14+6 + 5+19=56 \)
\( \bar{x}=\frac{56}{5}=11.2 \)

Step 2: Calculate the squared differences from the mean.

For each data point \( x_i \), calculate \( (x_i-\bar{x})^2 \):

• For \( x_1 = 12 \): \( (12 - 11.2)^2=(0.8)^2 = 0.64 \)
• For \( x_2 = 14 \): \( (14 - 11.2)^2=(2.8)^2 = 7.84 \)
• For \( x_3 = 6 \): \( (6 - 11.2)^2=(- 5.2)^2 = 27.04 \)
• For \( x_4 = 5 \): \( (5 - 11.2)^2=(-6.2)^2 = 38.44 \)
• For \( x_5 = 19 \): \( (19 - 11.2)^2=(7.8)^2 = 60.84 \)

Step 3: Calculate the sum of squared differences.

\( \sum_{i = 1}^{5}(x_i-\bar{x})^2=0.64 + 7.84+27.04 + 38.44+60.84 = 134.8 \)

Step 4: Calculate the sample variance ($s^2$) or population variance ($\sigma^2$). Since it's a sample (the problem says "a sample was done"), we use sample variance formula \( s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1} \)

\( s^2=\frac{134.8}{5 - 1}=\frac{134.8}{4}=33.7 \)

Step 5: Calculate the sample standard deviation ($s$)

\( s=\sqrt{s^2}=\sqrt{33.7}\approx5.8 \) (to one decimal place)

Answer:

\( 5.8 \)