Question

1. a student studying the sleeping habits of seniors at his school asked 36 randomly - selected seniors how many hours of sleep they got the previous night. the data, rounded to the nearest half - hour, is given in the table below.

8 7.5 9 7.5 9 6 5 9 7.5 7 8 7
6.5 8.5 8 6.5 8.5 6 7 7.5 7 6 8.5 7
7 8 7 7.5 7 6 7 8 7.5 6 7 5
a) using your calculator, find the mean and median of this data. without graphing it, what shape do you suspect, give the values of the mean and median?
b) find the five number summary for these data.
c) are there any outliers? justify your answer.
d) draw a modified boxplot of this distribution.
e) when looking at the boxplot of the distribution, how can you tell (without doing any calculations) that the mean of these data is larger than the median?
b) find and interpret the standard deviation.
c) suppose 4 more values were added to the data, each exactly equal to the mean. would this have any impact on the standard deviation? explain, without using any calculations.

Explanation:

Step1: Calculate the mean

Sum all the data values and divide by the number of data - points. There are $n = 36$ data - points. Let the data values be $x_1,x_2,\cdots,x_{36}$. The mean $\bar{x}=\frac{\sum_{i = 1}^{36}x_i}{36}$. Using a calculator, $\sum_{i=1}^{36}x_i=261$, so $\bar{x}=\frac{261}{36}=7.25$.

Step2: Calculate the median

First, order the data from smallest to largest. Since $n = 36$ (an even number), the median is the average of the $\frac{n}{2}=18$th and $(\frac{n}{2}+1)=19$th ordered data - values. After ordering the data: The 18th value is $7$ and the 19th value is $7.5$, so the median $M=\frac{7 + 7.5}{2}=7.25$.

Step3: Five - number summary

1. Minimum: The smallest value in the ordered data set is $5$.
2. First quartile ($Q_1$): Since $n = 36$, the position of $Q_1$ is at $\frac{n}{4}=9$th value in the ordered data set. The 9th value is $6.5$.
3. Median ($M$): We found it to be $7.25$.
4. Third quartile ($Q_3$): The position of $Q_3$ is at $\frac{3n}{4}=27$th value in the ordered data set. The 27th value is $8$.
5. Maximum: The largest value in the ordered data set is $9$.

Step4: Check for outliers

The inter - quartile range $IQR=Q_3 - Q_1=8 - 6.5 = 1.5$. The lower fence is $Q_1-1.5\times IQR=6.5-1.5\times1.5=6.5 - 2.25 = 4.25$. The upper fence is $Q_3 + 1.5\times IQR=8+1.5\times1.5=8 + 2.25 = 10.25$. Since all the data values are between $4.25$ and $10.25$, there are no outliers.

Step5: Standard deviation

Using a calculator, the sample standard deviation $s\approx1.04$. It represents the average amount by which each data value differs from the mean. A standard deviation of about $1.04$ indicates that the data values are relatively close to the mean.

Step6: Impact of adding values equal to the mean on standard deviation

Adding 4 values equal to the mean will decrease the standard deviation. The standard deviation measures the spread of the data. When we add values that are equal to the mean, the overall spread of the data decreases because these new values do not contribute to the deviation from the mean.

Answer:

a) Mean: $7.25$, Median: $7.25$
b) Five - number summary: Minimum = $5$, $Q_1 = 6.5$, Median = $7.25$, $Q_3 = 8$, Maximum = $9$
c) No outliers. The lower fence is $4.25$ and the upper fence is $10.25$, and all data values are within these fences.
d) (Drawing a box - plot is a visual task. The box extends from $Q_1 = 6.5$ to $Q_3 = 8$ with a line at the median $7.25$. The whiskers extend from the box to the minimum $5$ and maximum $9$).
e) Since the mean and median are equal, we cannot tell that the mean is larger than the median from the box - plot.
b) Standard deviation: $s\approx1.04$. It represents the average amount by which each data value differs from the mean.
c) Adding 4 values equal to the mean will decrease the standard deviation as these values do not contribute to the spread of the data.