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Q.100

Expert-verifiedFound in: Page 152

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$=1935$

$\mathrm{IQR}=245$

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

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) |

$2005-2006$ | $1585$ |

$2006-2007$ | $1690$ |

$2007-2008$ | $1735$ |

$2008-2009$ | $1935$ |

$2009-2010$ | $2021$ |

$2010-2011$ | $1890$ |

$\sum x=10856$ |

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

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

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

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 $=1935$

$\mathrm{IQR}=245$

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