Solved on Nov 27, 2023

Calculate the first quartile, third quartile, mean, median, range, standard deviation $s$ , interquartile range, lower and upper limits for outliers, and variance $s^2$ for the given GPA data of 15 students, rounded to 4 decimal places.

STEP 1

Assumptions1. The data represents the GPA of15 students. . The data is in ascending order.
3. We are using technology to calculate the statistics.

STEP 2

First, we need to sort the data in ascending order.
\begin{tabular}{|c|} \hline Sorted\, GPA \\ \hline1.52 \\ \hline1.62 \\ \hline1.62 \\ \hline1.83 \\ \hline1.92 \\ \hline2.68 \\ \hline2.74 \\ \hline2.9 \\ \hline2.95 \\ \hline.22 \\ \hline.26 \\ \hline.43 \\ \hline.63 \\ \hline.71 \\ \hline.97 \\ \hline\end{tabular}

STEP 3

(a) The first quartile (Q1) is the median of the lower half of the data. Since there are15 data points, the lower half is the first7 data points. The median of7 data points is theth data point.
$Q1 =1.83$

STEP 4

(b) The third quartile (Q3) is the median of the upper half of the data. The upper half is the last7 data points. The median of7 data points is the4th data point from the end.
$Q3 =3.43$

STEP 5

(c) The mean is the sum of all data points divided by the number of data points. $Mean = \frac{Sum\, of\, all\, data\, points}{Number\, of\, data\, points}$

STEP 6

Calculate the sum of all data points and the mean.
$Sum\, of\, all\, data\, points =1.52 +1.62 +1.62 +1.83 +1.92 +2.68 +2.74 +2.9 +2.95 +3.22 +3.26 +3.43 +3.63 +3.71 +3.97 =42.38$ $Mean = \frac{42.38}{15} =2.8253$

STEP 7

(d) The median is the middle value of the data. Since there are15 data points, the median is theth data point.
$Median =2.9$

STEP 8

(e) The range is the difference between the maximum and minimum data points.
$Range = Max\, data\, point - Min\, data\, point$ $Range =3.97 -1.52 =2.45$

STEP 9

(f) $s$ represents the standard deviation. It is calculated using the formula $s = \sqrt{\frac{\sum_{i=}^{n}(x_i - \bar{x})^2}{n-}}$ where $x_i$ are the data points, $\bar{x}$ is the mean, and $n$ is the number of data points.

STEP 10

Calculate the standard deviation.
$s = \sqrt{\frac{(.52-2.8253)^2 + (.62-2.8253)^2 + \ldots + (3.97-2.8253)^2}{15-}} =0.7414$

STEP 11

(g) The interquartile range (QR) is the difference between the third quartile and the first quartile.
$QR = Q3 - Q$ $QR =3.43 -.83 =.6$

STEP 12

(h) The lower limit (LL) and upper limit (UL) to find outliers are calculated using the formulas $LL = Q -.5 \times IQR$ $UL = Q +.5 \times IQR$

STEP 13

Calculate the lower limit and upper limit.
$LL =.83 -.5 \times.6 = -0.57$ $UL =3.43 +.5 \times.6 =5.83$

STEP 14

(i) $s^2$ represents the variance. It is the square of the standard deviation.
$s^2 = s^2 =0.7414^2 =0.5493$

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Calculate the first quartile, third quartile, mean, median, range, standard deviation $s$ , interquartile range, lower and upper limits for outliers, and variance $s^2$ for the given GPA data of 15 students, rounded to 4 decimal places.

STEP 1

Assumptions1. The data represents the GPA of15 students. . The data is in ascending order.
3. We are using technology to calculate the statistics.

STEP 2

STEP 3

(a) The first quartile (Q1) is the median of the lower half of the data. Since there are15 data points, the lower half is the first7 data points. The median of7 data points is theth data point.
$Q1 =1.83$

STEP 4

(b) The third quartile (Q3) is the median of the upper half of the data. The upper half is the last7 data points. The median of7 data points is the4th data point from the end.
$Q3 =3.43$

STEP 5

(c) The mean is the sum of all data points divided by the number of data points. $Mean = \frac{Sum\, of\, all\, data\, points}{Number\, of\, data\, points}$

STEP 6

Calculate the sum of all data points and the mean.
$Sum\, of\, all\, data\, points =1.52 +1.62 +1.62 +1.83 +1.92 +2.68 +2.74 +2.9 +2.95 +3.22 +3.26 +3.43 +3.63 +3.71 +3.97 =42.38$ $Mean = \frac{42.38}{15} =2.8253$

STEP 7

(d) The median is the middle value of the data. Since there are15 data points, the median is theth data point.
$Median =2.9$

STEP 8

(e) The range is the difference between the maximum and minimum data points.
$Range = Max\, data\, point - Min\, data\, point$ $Range =3.97 -1.52 =2.45$

STEP 9

(f) $s$ represents the standard deviation. It is calculated using the formula $s = \sqrt{\frac{\sum_{i=}^{n}(x_i - \bar{x})^2}{n-}}$ where $x_i$ are the data points, $\bar{x}$ is the mean, and $n$ is the number of data points.

STEP 10

Calculate the standard deviation.
$s = \sqrt{\frac{(.52-2.8253)^2 + (.62-2.8253)^2 + \ldots + (3.97-2.8253)^2}{15-}} =0.7414$

STEP 11

(g) The interquartile range (QR) is the difference between the third quartile and the first quartile.
$QR = Q3 - Q$ $QR =3.43 -.83 =.6$

STEP 12

(h) The lower limit (LL) and upper limit (UL) to find outliers are calculated using the formulas $LL = Q -.5 \times IQR$ $UL = Q +.5 \times IQR$

STEP 13

Calculate the lower limit and upper limit.
$LL =.83 -.5 \times.6 = -0.57$ $UL =3.43 +.5 \times.6 =5.83$

STEP 14

(i) $s^2$ represents the variance. It is the square of the standard deviation.
$s^2 = s^2 =0.7414^2 =0.5493$

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Calculate the first quartile, third quartile, mean, median, range, standard deviation sss, interquartile range, lower and upper limits for outliers, and variance s2s^2s2 for the given GPA data of 15 students, rounded to 4 decimal places.

STEP 1

Assumptions1. The data represents the GPA of15 students. . The data is in ascending order.3. We are using technology to calculate the statistics.

STEP 2

STEP 3

(a) The first quartile (Q1) is the median of the lower half of the data. Since there are15 data points, the lower half is the first7 data points. The median of7 data points is theth data point.Q1=1.83Q1 =1.83Q1=1.83

STEP 4

(b) The third quartile (Q3) is the median of the upper half of the data. The upper half is the last7 data points. The median of7 data points is the4th data point from the end.Q3=3.43Q3 =3.43Q3=3.43

STEP 5

(c) The mean is the sum of all data points divided by the number of data points.Mean=Sum of all data pointsNumber of data pointsMean = \frac{Sum\, of\, all\, data\, points}{Number\, of\, data\, points}Mean=NumberofdatapointsSumofalldatapoints​

STEP 6

STEP 7

(d) The median is the middle value of the data. Since there are15 data points, the median is theth data point.Median=2.9Median =2.9Median=2.9

STEP 8

(e) The range is the difference between the maximum and minimum data points.Range=Max data point−Min data pointRange = Max\, data\, point - Min\, data\, pointRange=Maxdatapoint−MindatapointRange=3.97−1.52=2.45Range =3.97 -1.52 =2.45Range=3.97−1.52=2.45

STEP 9

(f) sss represents the standard deviation. It is calculated using the formulas=∑i=n(xi−xˉ)2n−s = \sqrt{\frac{\sum_{i=}^{n}(x_i - \bar{x})^2}{n-}}s=n−∑i=n​(xi​−xˉ)2​​where xix_ixi​ are the data points, xˉ\bar{x}xˉ is the mean, and nnn is the number of data points.

STEP 10

Calculate the standard deviation.s=(.52−2.8253)2+(.62−2.8253)2+…+(3.97−2.8253)215−=0.7414s = \sqrt{\frac{(.52-2.8253)^2 + (.62-2.8253)^2 + \ldots + (3.97-2.8253)^2}{15-}} =0.7414s=15−(.52−2.8253)2+(.62−2.8253)2+…+(3.97−2.8253)2​​=0.7414

STEP 11

(g) The interquartile range (QR) is the difference between the third quartile and the first quartile.QR=Q3−QQR = Q3 - QQR=Q3−QQR=3.43−.83=.6QR =3.43 -.83 =.6QR=3.43−.83=.6

STEP 12

(h) The lower limit (LL) and upper limit (UL) to find outliers are calculated using the formulasLL=Q−.5×IQRLL = Q -.5 \times IQRLL=Q−.5×IQRUL=Q+.5×IQRUL = Q +.5 \times IQRUL=Q+.5×IQR

STEP 13

Calculate the lower limit and upper limit.LL=.83−.5×.6=−0.57LL =.83 -.5 \times.6 = -0.57LL=.83−.5×.6=−0.57UL=3.43+.5×.6=5.83UL =3.43 +.5 \times.6 =5.83UL=3.43+.5×.6=5.83

STEP 14

(i) s2s^2s2 represents the variance. It is the square of the standard deviation.s2=s2=0.74142=0.5493s^2 = s^2 =0.7414^2 =0.5493s2=s2=0.74142=0.5493

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Calculate the first quartile, third quartile, mean, median, range, standard deviation $s$ , interquartile range, lower and upper limits for outliers, and variance $s^2$ for the given GPA data of 15 students, rounded to 4 decimal places.

Assumptions1. The data represents the GPA of15 students. . The data is in ascending order.
3. We are using technology to calculate the statistics.

(a) The first quartile (Q1) is the median of the lower half of the data. Since there are15 data points, the lower half is the first7 data points. The median of7 data points is theth data point.
$Q1 =1.83$

(b) The third quartile (Q3) is the median of the upper half of the data. The upper half is the last7 data points. The median of7 data points is the4th data point from the end.
$Q3 =3.43$

(c) The mean is the sum of all data points divided by the number of data points. $Mean = \frac{Sum\, of\, all\, data\, points}{Number\, of\, data\, points}$

(d) The median is the middle value of the data. Since there are15 data points, the median is theth data point.
$Median =2.9$

(e) The range is the difference between the maximum and minimum data points.
$Range = Max\, data\, point - Min\, data\, point$ $Range =3.97 -1.52 =2.45$

(f) $s$ represents the standard deviation. It is calculated using the formula $s = \sqrt{\frac{\sum_{i=}^{n}(x_i - \bar{x})^2}{n-}}$ where $x_i$ are the data points, $\bar{x}$ is the mean, and $n$ is the number of data points.

Calculate the standard deviation.
$s = \sqrt{\frac{(.52-2.8253)^2 + (.62-2.8253)^2 + \ldots + (3.97-2.8253)^2}{15-}} =0.7414$

(g) The interquartile range (QR) is the difference between the third quartile and the first quartile.
$QR = Q3 - Q$ $QR =3.43 -.83 =.6$

(h) The lower limit (LL) and upper limit (UL) to find outliers are calculated using the formulas $LL = Q -.5 \times IQR$ $UL = Q +.5 \times IQR$

Calculate the lower limit and upper limit.
$LL =.83 -.5 \times.6 = -0.57$ $UL =3.43 +.5 \times.6 =5.83$

(i) $s^2$ represents the variance. It is the square of the standard deviation.
$s^2 = s^2 =0.7414^2 =0.5493$