Algebra II - Standard Deviation

A jellybean machine dispenses 3 jellybeans on the first trial, 5 jellybeans on the second trial, and 7 jellybeans on the third trial. Determine the sample standard deviation.

Explanation

The standard deviation measures the spread of the results.

Write the formula for the sample standard deviation.

$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\overline{x})^2}$

Determine the mean $\overline{x}$ .

$\overline{x} = \frac{3+5+7}{3} = \frac{15}{3} =5$

The term $\sum_{i=1}^{N}(x_i-\overline{x})^2$ means that we are summing the squared differences from the mean.

$\sum_{i=1}^{N}(x_i-\overline{x})^2 = (3-5)^2+(5-5)^2+(7-5)^2$

Simplify the terms.

$\sum_{i=1}^{N}(x_i-\overline{x})^2= (-2)^2+0+(2)^2 = 4+4 = 8$

represents the sample size. There are three numbers in the data.

$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\overline{x})^2} = \sqrt{\frac{1}{3-1}(8)}=\sqrt{4}=2$

The answer is:

Mr. Bell gave out a science test last week to six honors students. The scores were 88, 94, 80, 79, 74, and 83. What is the standard deviation of the scores? (Round to the nearest tenth.)

Explanation

First, find the mean of the six numbers by adding them all together, and dividing them by six.

88 + 94 + 80 + 79 + 74 + 83 = 498

498/6 = 83

Next, find the variance by subtracting the mean from each of the given numbers and then squaring the answers.

88 – 83 = 5

52 = 25

94 – 83 = 11

112 = 121

80 – 83 = –3

–32 = 9

79 – 83 = –4

–42 = 16

74 – 83 = –9

–92 = 81

83 – 83 = 0

02 = 0

Find the average of the squared answers by adding up all of the squared answers and dividing by six.

25 + 121 + 9 +16 +81 + 0 = 252

252/6 = 42

42 is the variance.

To find the standard deviation, take the square root of the variance.

The square root of 42 is 6.481.

In her last six basketball games, Jane scored 15, 17, 12, 15, 18, and 22 points per game. What is the standard deviation of these score totals? Round your answer to the nearest tenth.

Explanation

The following is the formula for standard deviation:

$s = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2}$

Here is a breakdown of what that formula is telling you to do:

1. Solve for the mean (average) of the five test scores
2. Subtract that mean from each of the five original test scores. Square each of the differences.
3. Find the mean (average) of each of these differences you found in Step 2
4. Take the square root of this final mean from #3. This is the standard deviation

Here are those steps:

1. Find the mean of her score totals:

$(15+17+12+15+18+22) \div 6 = 99 \div 6 =16.5$

2. Subtract the mean from each of the test scores, then square the differences:

3. Find the mean of the squared values from Step 2:

$(2.25+0.25+20.25+2.25+2.25+30.25) \div6 =$

$57.5\div 6= 9.58\bar{3}$

4. Take the square root of your answer from Step 3:

$\sqrt{9.58\bar{3}} = 3.096 = 3.1$

Determine the population standard deviation of the following data set and round to three decimal places:

Explanation

Write the formula for population standard deviation.

$\sigma= \sqrt{\frac{1}{N}\sum_{i}^{N}(x_i-\mu)^2}$

represents the number of terms, represents the terms in the data set, and $\mu$ is the mean.

Calculate the mean, $\mu$ .

$\frac{1+9-7}{3} = \frac{3}{3}=1$

$\mu =1$

Evaluate the variance, $\frac{1}{N}\sum_{i}^{N}(x_i-\mu)^2$ .

$\frac{1}{3}[(1-1)^2+(9-1)^2+(-7-1)^2] = \frac{1}{3}[64+64]=\frac{128}{3}$

The standard deviation is the square root of the variance.

$\sigma= \sqrt{\frac{1}{N}\sum_{i}^{N}(x_i-\mu)^2} = \sqrt{\frac{128}{3}}= 6.532$

The answer is:

If standard deviation is and the mean is , what is the range of the number set if it's within one standard deviation?

Explanation

Standard deviation is the dispersion of the data set. Since it's asking for within one standard deviation, we need to take the mean and add the standard deviation to find the upper bound of the range. Then, we will need to subtract the standard deviation from the mean to identify the lower bound of the range.

If the mean is $\small 5$ with a standard deviation of $\small 2$ , then which of the following values is within one standard deviation?

$\small 4$

$\small 8$

$\small 2.999999$

$\small 7.000001$

$\small 1$

Explanation

If mean is $\small 5$ with standard deviation of $\small 2$ , then one standard deviation within has a range of $\small \small 3$ to $\small 7$ .

Remember, we find the range by adding the standard deviation to the mean and subtracting the standard deviation from the mean.

$(5-2,5+2)\rightarrow (3,7)$

Only $\small 4$ is in the range.

The rest of the numbers are more than one standard deviation.

What is the difference between two standard deviations on the right tail with one standard deviation on the right tail? Assume a normal distribution.

$\small 13.5$

$\small 27$

$\small 15$

$\small 47.5$

$\small 34$

Explanation

Two standard deviations represents $\small 95%$ . One standard deviation represents $\small 68%$ . The difference is $\small 27%$ . However, the question is focusing on the right side of the tail. Since it's normal distribution, both tails of the graph are equal. Divide $\small 27$ by $\small 2$ and we get $\small 13.5$ .

At the end of the fall semester, a math class of ninth graders had the following grades: 85, 75, 97, 83, 62, 75, 88, 84, 92, and 89.

What is the standard deviation of this class?

Explanation

The standard deviation of a set of numbers is how much the numbers deviate from the mean. More formally, the standard deviation is

$\sigma =\sqrt{\frac{\sum{(x-M)^2} }{N-1}}$

where is a number in the series, is the mean, and is the number of data points. So, to calculate the standard deviation, we must first calculate the mean. The mean of this data set is

$M = \frac{85+75+97+83+62+75+88+84+92+89}{10}=83$

Now that we know the mean, we can start calculating the standard deviation. We first need to find the sum of each data point minus the average squared.

Calculating that, we get that the variance from the mean is . Plugging that into our equation for standard deviation, with being ten data points, we get

$\sigma = \sqrt{\frac{912}{9}}=10.07$

Determine the sample standard deviation if the sample variance is .

Explanation

Write the formula for the standard deviation given the variance.

$\textup{Standard Deviation }(\sigma) = \sqrt{\textup{Variance}}$

Substitute the variance into the standard deviation.

$\sigma=\sqrt{225} = 15$

The answer is:

In the population of high school boys, the variance in height, measured in inches, was found to be 16. Assuming that the height data is normally distributed, 95% of high school boys should have a height within how many inches of the mean?

$8\ \textup{inches}$

$4\ \textup{inches}$

$\textup{Cannot be determined from the information given}$

$16\ \textup{inches}$

$2\ \textup{inches}$

Explanation

The 68-95-99.7 rule states that nearly all values lie within 3 standard deviations of the mean in a normal distribution. In this case the question asks for 95% so we want to know what 2 standard deviations from the mean is.

We are given the variance, so to find the standard deviation, take the square root.

$\sqrt{16}=4$

So two standard devations is 8 inches. 95% of heights should be within 8 inches of the mean.

Standard Deviation

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