Question

question 4
4.1 the diagram below shows the recorded monthly rainfall statistics for october in 2004 to 2015.
5 32 49 103 197
4.1.1 determine the range of the rainfall for the october months.
4.1.2 determine the semi - interquartile range of the rainfall for the october months.
4.1.3 comment on the skewness of the data for the october months.
4.1.4 if the maximum rainfall was incorrectly recorded and is increased, what effect will this have on:
(a) the mean and
(b) median
4.1.5 for how many years was october’s rainfall less than 32 mm?
4.2 on the 24th of september 2024, a tour operator sent 11 tour buses to 11 destinations at 9am. the table below shows the number of passengers
8 8 10 12 16 19 20 21 24 25
4.2.1 calculate the standard deviation for this data set
4.2.2 a tour is regarded as popular if the number of passengers on a tour is one standard deviation above the mean.
how many destinations were popular on this day?
4.2.3 on the 24th of september 2024, an additional bus was sent to a destination at 10am. the number of passengers was unknown for all 12 buses was 19.
8 8 10 12 16 19 20 21 24
determine x, the number of passengers in the 12th bus.

Explanation:

4.1.1

Find the difference between max and min.
$197 - 5=192$

4.1.2

First, find the quartiles. The data set has 12 values. The median (second - quartile $Q_2$) is the average of the 6th and 7th ordered values. The lower half of the data is the first 6 values, and its median (first - quartile $Q_1$) is the average of the 3rd and 4th ordered values. The upper half of the data is the last 6 values, and its median (third - quartile $Q_3$) is the average of the 9th and 10th ordered values.
Ordered data: 5, 32, 49, 103, 197
$Q_1=\frac{32 + 49}{2}=40.5$
$Q_3=\frac{103+197}{2}=150$
Semi - interquartile range $=\frac{Q_3 - Q_1}{2}=\frac{150 - 40.5}{2}=54.75$

4.1.3

Since the distance from $Q_1$ to the median is much smaller than the distance from the median to $Q_3$, the data is positively skewed.

4.1.4

(a)

The mean is $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. If the maximum value is increased, the sum $\sum_{i = 1}^{n}x_i$ increases while $n$ remains the same. So the mean will increase.

(b)

The median is the middle - value (or average of two middle - values) when the data is ordered. Increasing the maximum value does not change the position of the middle - value(s) for a fixed number of data points. So the median will remain the same.

4.1.5

From the box - and - whisker plot, the value 32 is the lower quartile. Counting the number of years with rainfall less than 32 mm, we assume that the number of years is 3 (by looking at the part of the data before the lower quartile).

4.2.1

1. First, find the mean $\bar{x}$ of the data set $\{8,8,10,12,16,19,20,21,24,25\}$.

$\bar{x}=\frac{8 + 8+10+12+16+19+20+21+24+25}{10}=\frac{163}{10}=16.3$

1. Then, find the squared differences $(x_i-\bar{x})^2$ for each data point:

$(8 - 16.3)^2=(- 8.3)^2 = 68.89$ (twice)
$(10 - 16.3)^2=(-6.3)^2 = 39.69$
$(12 - 16.3)^2=(-4.3)^2 = 18.49$
$(16 - 16.3)^2=(-0.3)^2 = 0.09$
$(19 - 16.3)^2=(2.7)^2 = 7.29$
$(20 - 16.3)^2=(3.7)^2 = 13.69$
$(21 - 16.3)^2=(4.7)^2 = 22.09$
$(24 - 16.3)^2=(7.7)^2 = 59.29$
$(25 - 16.3)^2=(8.7)^2 = 75.69$

1. Next, find the variance $s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}$

$\sum_{i = 1}^{10}(x_i-\bar{x})^2=2\times68.89+39.69+18.49+0.09+7.29+13.69+22.09+59.29+75.69=334.2$
$s^2=\frac{334.2}{9}\approx37.13$

1. Finally, the standard deviation $s=\sqrt{s^2}=\sqrt{37.13}\approx6.1$

4.2.2

The mean $\bar{x}=16.3$ and the standard deviation $s\approx6.1$. One standard deviation above the mean is $16.3+6.1 = 22.4$. Counting the number of values in the data set greater than 22.4, we have 2 (24 and 25).

4.2.3

The mean of the 12 - bus data is 19. The sum of the number of passengers in 11 buses is $8 + 8+10+12+16+19+20+21+24+25+x$.
The sum of the 11 known values is $8+8+10+12+16+19+20+21+24+25 = 143$.
We know that the mean of 12 values is 19, so the sum of 12 values is $19\times12 = 228$.
Then $143+x=228$, so $x=228 - 143=85$

Answer:

4.1.1: 192
4.1.2: 54.75
4.1.3: Positively skewed
4.1.4: (a) Increase; (b) Remain the same
4.1.5: 3
4.2.1: Approximately 6.1
4.2.2: 2
4.2.3: 85