Math

QuestionCalculate the standard deviation for the data: 1 | 877868, 2 | 0323.

Studdy Solution

STEP 1

Assumptions1. The data is given in a stem-and-leaf plot. . The stem represents the tens place and the leaves represent the ones place.
3. The data is ungrouped, meaning each leaf represents a single data point.

STEP 2

First, we need to convert the stem-and-leaf plot into a list of data points. This is done by combining each stem with each of its leaves.
The data points are18,17,18,16,18,20,20,23,23.

STEP 3

To find the standard deviation, we first need to find the mean (average) of the data. The mean is found by summing all the data points and dividing by the number of data points.
Mean=SumofdatapointsNumberofdatapointsMean = \frac{Sum\, of\, data\, points}{Number\, of\, data\, points}

STEP 4

Plug in the values for the sum of data points and the number of data points to calculate the mean.
Mean=18+17+18+16+18+20+20+23+239Mean = \frac{18+17+18+16+18+20+20+23+23}{9}

STEP 5

Calculate the mean.
Mean=173919.22Mean = \frac{173}{9} \approx19.22

STEP 6

Next, we need to calculate the variance. The variance is the average of the squared differences from the mean.
Variance=Sumof(datapointMean)2NumberofdatapointsVariance = \frac{Sum\, of\, (data\, point - Mean)^2}{Number\, of\, data\, points}

STEP 7

Subtract the mean from each data point, square the result, and sum all these squared differences.
Variance=(1819.22)2+(1719.22)2+(1819.22)2+(1619.22)2+(1819.22)2+(2019.22)2+(2019.22)2+(2319.22)2+(2319.22)29Variance = \frac{(18-19.22)^2+(17-19.22)^2+(18-19.22)^2+(16-19.22)^2+(18-19.22)^2+(20-19.22)^2+(20-19.22)^2+(23-19.22)^2+(23-19.22)^2}{9}

STEP 8

Calculate the variance.
Variance4.20Variance \approx4.20

STEP 9

Finally, the standard deviation is the square root of the variance.
StandardDeviation=VarianceStandard\, Deviation = \sqrt{Variance}

STEP 10

Plug in the value for the variance to calculate the standard deviation.
StandardDeviation=4.20Standard\, Deviation = \sqrt{4.20}

STEP 11

Calculate the standard deviation.
StandardDeviation.05Standard\, Deviation \approx.05The standard deviation for the group of data items is approximately.05.

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