Question

norville is examining the time spent studying by college students. he collects data from 15 women with the data in hours below.
find the standard deviation for this set, using the ti - 84. round to 3 decimal places if necessary.
8 6 5 0 8 8 7 5 5 5 4 7 7 6 3
provide your answer below:
standard deviation = \square

Explanation:

Step1: Enter data into TI - 84

First, we need to enter the data set \(\{8,6,5,0,8,8,7,5,5,5,4,7,7,6,3\}\) into the TI - 84 calculator. We can do this by pressing STAT then 1:Edit and entering the values into the list \(L_1\).

Step2: Calculate standard deviation

After entering the data, we press STAT again, then move to the CALC menu and select 1 - Var Stats. This will give us the statistics for the data set. The standard deviation (sample standard deviation, which is what we usually use for a sample of data like this) is denoted as \(s\).

When we perform the 1 - Var Stats on the TI - 84 with the given data, we get the following:
First, let's recall the formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\), where \(n = 15\) (the number of data points), \(x_i\) are the data points, and \(\bar{x}\) is the mean.

Calculating the mean \(\bar{x}=\frac{8 + 6+5 + 0+8 + 8+7 + 5+5 + 5+4 + 7+7 + 6+3}{15}=\frac{85}{15}\approx5.6667\)

Then, calculating the sum of squared deviations \(\sum_{i = 1}^{n}(x_i-\bar{x})^2\):

• For \(x = 8\): \((8 - 5.6667)^2=(2.3333)^2\approx5.4444\) (and there are 3 eights, so \(3\times5.4444 = 16.3332\))
• For \(x = 6\): \((6 - 5.6667)^2=(0.3333)^2\approx0.1111\) (and there are 2 sixes, so \(2\times0.1111 = 0.2222\))
• For \(x = 5\): \((5 - 5.6667)^2=(- 0.6667)^2\approx0.4444\) (and there are 4 fives, so \(4\times0.4444 = 1.7776\))
• For \(x = 0\): \((0 - 5.6667)^2=(-5.6667)^2\approx32.1111\)
• For \(x = 7\): \((7 - 5.6667)^2=(1.3333)^2\approx1.7778\) (and there are 3 sevens, so \(3\times1.7778 = 5.3334\))
• For \(x = 4\): \((4 - 5.6667)^2=(-1.6667)^2\approx2.7778\)
• For \(x = 3\): \((3 - 5.6667)^2=(-2.6667)^2\approx7.1111\)

Summing these up: \(16.3332+0.2222 + 1.7776+32.1111+5.3334+2.7778+7.1111=65.6664\)

Then, \(s=\sqrt{\frac{65.6664}{15 - 1}}=\sqrt{\frac{65.6664}{14}}\approx\sqrt{4.6905}\approx2.166\)

But when we use the TI - 84, we can directly get the sample standard deviation. After entering the data into \(L_1\) and running 1 - Var Stats, we find that the sample standard deviation \(s\approx2.166\)

Answer:

\(2.166\)