91影视

Given IQ score:

$145,139,126,122,125,130,96,110,118,118$

For option (a), order the data values from smallest to largest:

$96,110,118,118,122,125,126,130,139,145$

The median is the middle value of the data. And since the number of data is even then:

$$\begin{gathered} M=Q_2=\frac{122+125}{2} \\ =\frac{247}{2} \\ =123.5 \end{gathered}$$

The first quartile is the median of the data values below the median i.e.

$Q_1=118$

The third quartile is the median of the data values above the median i.e.

$Q_3=130$

Now, the interquartile range will be:

$$\begin{gathered} IQR=Q_3-Q_1 \\ =130-118 \\ =12 \end{gathered}$$

Outliers are observations that are more than $1.5$times the IQR above $Q_{3}orbelow{Q}_1.$

$$\begin{gathered} Q_3+1.5IQR=130+1.5\left(12\right)=148 \\ {Q}_1-1.5IQR=118-1.5\left(12\right)=100 \end{gathered}$$

Since$96$ lies below $100$,$96$ is considered an outlier by the$1.5$ times interquartile range and thus the option (a) is true.

For option(b), the five summary consists of the minimum, first quartile, median, third quartile and the maximum, which are all the same thus the option (b) is true.

For option (c), $96$is the smallest value in the data set. We then expect the mean to increase if$96$ is removed from the data set, as the centre of the data will become higher when the minimum is removed and thus, option (c) is true.

For option (d), the standard deviation measures the amount of variation or the spread of the data set. $96$is the smallest value in the data set. We then expect the standard deviation to increase if$96$ is removed from the data set, as the spread decrease when the minimum is removed and thus, option (d) is true.

For option (e), $96$is the smallest value in the data set and from part (a) we know that$96$ is an outlier. We then expect the IQR to not be affected if $96$is removed from the data as the IQR should not be influenced by the outlier and thus option (e) is not true.

Thus, we will select the option (e) as correct answer.