Question

the mean and median are listed for a distribution. determine the shape of the distribution. justify your answer.

1. paved line to work: mean = 63.28 and median = 42.5
2. age of students: mean = 16.87 and median = 17
3. reaction time: mean = 0.451 and median = 0.376
4. number of texts sent in a day: mean = 11 and median = 15
5. grade level: mean = 11 and median = 11
6. hours watching tv in a week: mean = 15.75 and median = 8
7. students were asked to measure their arm - span in centimeters. a random sample of 18 students yielded the results below.

armspan (cm)
60 166
76 173
113 176
148 175
156 177
160 177
160 181
160 183
160 185
a.) calculate the mean and the median of the distribution.
b.) based on the part (a) what is the shape of the distribution?
c.) graph the box - plot
d.) graph the histogram, starting at 60 with a bin width of 10.
e.) now that you have graphed the distribution, is your answer to part b the same?

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. First, sum up all the data - points: $60+76 + 113+148+156+160+160+160+173+176+175+177+177+181+183+183+185+186=2949$. There are $n = 18$ data - points. So, $\bar{x}=\frac{2949}{18}\approx163.83$.

Step2: Calculate the median

Arrange the data in ascending order: $60,76,113,148,156,160,160,160,173,175,176,177,177,181,183,183,185,186$. Since $n = 18$ (an even number), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered data - points. $\frac{n}{2}=9$ and $\frac{n}{2}+1 = 10$. The 9th value is $173$ and the 10th value is $175$. So, the median $M=\frac{173 + 175}{2}=174$.

Step3: Determine the shape of the distribution

Since the mean ($\bar{x}\approx163.83$) is less than the median ($M = 174$), the distribution is left - skewed. In a left - skewed distribution, the tail of the distribution extends to the left, pulling the mean in that direction.

Step4: Graph the box - plot

1. First, find the five - number summary:

• Minimum: $60$
• First quartile ($Q_1$): Since $n = 18$, the position of $Q_1$ is $\frac{n+1}{4}=4.75$. So, $Q_1$ is $0.25$ of the way between the 4th and 5th ordered data - points. The 4th value is $148$ and the 5th value is $156$. $Q_1=148+(156 - 148)\times0.25=150$.
• Median ($Q_2$): $174$
• Third quartile ($Q_3$): The position of $Q_3$ is $\frac{3(n + 1)}{4}=14.25$. So, $Q_3$ is $0.25$ of the way between the 14th and 15th ordered data - points. The 14th value is $181$ and the 15th value is $183$. $Q_3=181+(183 - 181)\times0.25=181.5$.
• Maximum: $186$

1. Draw a number line that includes the range from the minimum to the maximum. Draw a box from $Q_1$ to $Q_3$, with a vertical line at the median. Draw whiskers from the box to the minimum and maximum values.

Step5: Graph the histogram

The bins start at $60$ with a bin width of $10$. The bins are: $60 - 69,70 - 79,80 - 89,90 - 99,100 - 109,110 - 119,120 - 129,130 - 139,140 - 149,150 - 159,160 - 169,170 - 179,180 - 189$. Count the number of data - points in each bin:

Bin	Frequency
$70 - 79$	$1$
$80 - 89$	$0$
$90 - 99$	$0$
$100 - 109$	$0$
$110 - 119$	$1$
$120 - 129$	$0$
$130 - 139$	$0$
$140 - 149$	$1$
$150 - 159$	$1$
$160 - 169$	$4$
$170 - 179$	$5$
$180 - 189$	$4$

Draw rectangles for each bin with the height equal to the frequency.

Step6: Re - evaluate the shape

Looking at the histogram and box - plot, the left - skewness is still evident. The left - hand side of the histogram has fewer and smaller bars, and the box - plot shows the left whisker is longer than the right whisker.

Answer:

a. Mean $\approx163.83$, Median $ = 174$
b. Left - skewed
c. Box - plot is drawn with minimum $=60$, $Q_1 = 150$, Median $=174$, $Q_3 = 181.5$, maximum $=186$
d. Histogram is drawn with bins and frequencies as shown above
e. Yes, the answer to part b is still left - skewed.