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Q.100

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Found in: Page 152

### Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208

# Use the following information to answer the next nine exercises: The population parameters below describe the full-time equivalent number of students (FTES) each year at Lake Tahoe Community College from 1976–1977 through 2004–2005. • μ = 1000 FTES • median = 1,014 FTES • σ = 474 FTES • first quartile = 528.5 FTES • third quartile = 1,447.5 FTES• n = 29 years Calculate the mean, median, standard deviation, the first quartile, the third quartile and the IQR. Round to one decimal place

Mean $=1809.3$

Median $=1812.5$

Standard deviation $=151.2$

First quartile $=1690$

Third quartile$=1935$

$\mathrm{IQR}=245$

See the step by step solution

## Step 1 : To find

The mean, median, standard deviation, the first quartile, the third quartile and the IQR.

## Step 2: To calculating the mean value.

The mean is one of the measurements of central tendency in statistics. The mean is defined as the total of data values divided by the number of data values. The following is the formula for determining the mean:

Mean $=\mu =\frac{\sum x}{N}$

Calculation: We are given below the data.

 YEAR TOTAL FTES (x) $2005-2006$ $1585$ $2006-2007$ $1690$ $2007-2008$ $1735$ $2008-2009$ $1935$ $2009-2010$ $2021$ $2010-2011$ $1890$ $\sum x=10856$

$Mean=\frac{10856}{6}=1809.3$

## Step 3: To calculating median value

First arrange the data in the ascending order

$1585,1690,1735,1890,1935,2021$

Since we have even number of observations, therefore, the median is:

Median $=\frac{1735+1890}{2}=1812.5$

Hence the median value is = 1812.5

## Step 4: To calculate the standard deviation value.

$\begin{array}{rl}\text{Standard deviation}& =\sqrt{\frac{\left(1585-1809.3{\right)}^{2}+\left(1690-1809.3{\right)}^{2}+\dots +\left(1890-1809.3{\right)}^{2}}{6}}\\ SD& =\sqrt{\frac{50310.49+14232.49+\dots +137193.3}{6}}\\ SD& =151.2\end{array}$

## Step 5: To find first and third quartile values and IQR values

First we have to arrange the data in ascending order.

$1585,1690,1735,1890,1935,2021$

${Q}_{1}=1690$ because it is middle term of the first $50%$of the data

${Q}_{3}=1935$ because it the middle term of the last $50%$ of the data

Now, the interquartile range is calculated as:

$IQR={Q}_{3}-{Q}_{1}=1935-1690=245$

Mean $=1809.3$

Median $=1812.5$

Standard deviation =$=151.2$$=151.2$

First quartile $=1690.$

Third quartile $=1935$

$\mathrm{IQR}=245$