Question

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

Short answer

Mean$=1809.3$

Median $=1812.5$

Standard deviation$=151.2$

First quartile$=1690$

Third quartile$=1935$

$\mathrm{IQR}=245$

Step-by-step solution

To find

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

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

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

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

To calculate the standard deviation value.

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

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:

$IQR=Q_3-Q_1=1935-1690=245$

Final answer

Mean$=1809.3$

Median$=1812.5$

Standard deviation =$=151.2$$=151.2$

First quartile$=1690.$

Third quartile$=1935$

$\mathrm{IQR}=245$