Algebra - How to find standard deviation

In meteorology, the standard deviation of wind speed can be used to predict the likelihood of fog forming under given temperature conditions.

What is the standard deviation of the following wind speed measurements in kilometers per hour (kph), taken 1 hour apart at the same site for 10 hours? Round to the nearest tenth.

Explanation

The first step in calculating standard deviation, or $\sigma$ , is to calculate the mean for your sample, or $\mu$ . Remember, to calculate mean, sum your data values and divide by the count, or number of values you have.

$\mu= \frac{x_{1} +x_{2} + ...x_{n}}{n} = \frac{6+4+13+11+8+6+4+5+7+7}{10} = 7.1$

Next, we must find the difference between each recorded value and the mean. At the same time, we will square these differences, so it does not matter whether you subtract the mean from the value or vice versa.

We use $\sum_{i=1}^{N} (x_{i}-\mu)^{2}$ to represent this, but all it really means is that you square the difference between each value $x_{i}$ , where is the position of the value you're working with, and the mean, $\mu$ . Then we sum all those differences up (the part that goes $\sum_{i=1}^{N}$ , where is your count. just refers to the fact that you start at the first value, so you include them all.)

It's probably easier to do than to think about at first, so let's dive in!

$(6-7.1)^{2} = 1.21$

$(4-7.1)^{2} = 9.61$

$(13-7.1)^{2} = 34.81$

$(11-7.1)^{2} = 15.21$

$(8-7.1)^{2} = .81$

$(6-7.1)^{2} = 1.21$

$(4-7.1)^{2} = 9.61$

$(5-7.1)^{2} = 4.41$

$(7-7.1)^{2} = .01$

$(10-7.1)^{2} = 8.41$

Now, add the deviations, and we're nearly there!

$\sum_{i=1}^{N} (x_{i}-\mu)^{2} = 1.21 + 6.91 + 34.81 + ... + 8.41 = 85.29$

Next, we must divide this number by our :

$\frac{85.29}{N} = 8.529$

This number, 8.529, is our variance, or $\sigma^{2}$ . Since standard variation is $\sigma$ , you may have guessed what we must do next. We must take the square root of the summed squares of deviations.

$\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N} (x_{i}-\mu)^{2}} = \sqrt{8.529} = 2.92$

So, our standard deviation is 2.9 kph (remembering the problem told us to round to 1 decimal point.)

Approximate the standard deviation of the given set of numbers to three decimal places:

Explanation

Write the formula for standard deviation.

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

represents the total numbers in the data set:

$\mu$ is the mean: $\mu = \frac{1+2+9}{3} =\frac{12}{3} = 4$

Find the mean of the squared differences. Subtract each number in the data set with the mean, square the quantities, and take the average.

This will be the variance:

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

$\frac{1}{3} \cdot [ (9-4)^2+(2-4)^2+(1-4)^2]$

$\frac{1}{3} \cdot [ (5)^2+(-2)^2+(-3)^2] = \frac{1}{3}[25+4+9]= \frac{38}{3}$

The standard deviation is the square root of the variance, which is the square root of this fraction.

$\sqrt{\frac{38}{3}} = 3.559$

The answer is:

Find the standard deviation of the following set of numbers:

$\sqrt{6}$

$\sqrt{7}$

$\sqrt{10}$

Explanation

To find standard deviation, we will follow these steps:

Find the mean of the original set of numbers.
Subtract the mean from each of the original numbers.
Square each number.
Find the mean of the new set of numbers.
Take the square root of the mean.

So, we will take this step by step.

STEP 1

$\text{mean} = \frac{7 + 10 + 13}{3}$

$\text{mean} = \frac{30}{3}$

$\text{mean} = 10$

STEP 2

7, 10, 13

(7-10), (10-10), (13-10)

-3, 0, 3

STEP 3

9, 0, 9

STEP 4

$\text{mean} = \frac{9+0+9}{3}$

$\text{mean} = \frac{18}{3}$

$\text{mean} = 6$

STEP 5

$\sqrt{6}$

Thefore, the standard deviation is $\sqrt{6}$ .

Give the interquartile range of a data set with the following characteristics.

Mean: 72.1

Median: 70

Standard deviation: 4.6

It cannot be determined from the information given.

Explanation

The interquartile range is the difference between the first and third quartiles. The two pieces of information needed to determine interquartile range, the first and third quartiles, are missing; therefore, it is impossible to answer the question without more information.

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.

Find the standard deviation. Round your answer to the nearest tenth.

Explanation

To find standard deviation, we apply this formula: $\sqrt{\frac{1}{n}\Sigma (\bar{x}-x)^2}$ .

n represents how many data in the set.

$\bar{x}$ represents the average of the data in the set.

is any data in the set.

$\Sigma$ is summation of the difference between average and any data value squared.

So the mean is $\frac{1+3+5+7+9}{5}=5$ . Now, we apply the formula.

$\sqrt{\frac{1}{5}[(5-1)^2+(5-3)^2+(5-5)^2+(5-7)^2+(5-9)^2]}}$

$=\sqrt{\frac{1}{5}[16+4+0+4+16])}$

$=\sqrt{\frac{1}{5}(40)}$

Find the standard deviation of the data set:

Explanation

Write the formula of standard deviation.

$\sigma=\sqrt{\frac{1}{N}\sum_{a=1}^{N}(x_a-\mu)^2}= \sqrt{\textup{Variance}}$

Determine the mean, $\mu$ .

$\mu=\frac{1+3+8}{3}= \frac{12}{3}= 4$

Subtract each number in the dataset from the mean and square each result.

This is the $x_a-\mu$ term.

Find the mean of the squared differences, or the variance.

This is the $\frac{1}{N}\sum_{a=1}^{N}(x_a-\mu)^2$ term.

$\textup{Variance}= \frac{9+1+16}{3}=\frac{26}{3}$

Square root the variance for the standard deviation.

$\sigma=\sqrt{\frac{26}{3}}= 2.944$

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:

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