91Ó°ÊÓ

Find the median, quartiles, and interquartile range:

Here there are $8$frequencies listed out

$1,5,10,15,18,22,24,24$

Therefore, there are eight numbers, so, the median for even numbers is that average of two middle numbers.

$Median=\frac{15+18}{2}=16.5$

The first quartile is the middle point of $3$frequencies below the Median.

$Q_1=5$

The third quartile is the middle point of $3$frequencies above the Median.

$Q_3=24$

Calculate IQR

$\mathrm{IQR}={\mathrm{Q}}_3−{\mathrm{Q}}_1=24−5=19$

Calculate $1.5â‹\dots \mathrm{IQR}$below the first quartile and check for low outliers.

role="math" localid="1650200003843" ${\mathrm{Q}}_1−1.5â‹\dots \mathrm{IQR}=5-1.5×19=-23.5$

There are no low outliers.

Calculate $1.5â‹\dots \mathrm{IQR}$above the third quartile and check for high outliers.

${\mathrm{Q}}_3+1.5â‹\dots \mathrm{IQR}=24+1.5×19=52.5$

There are no high outliers.

Conclusion: there are no low and high outliers in the given data sets of frequencies of the book publisher.

If outliers are there, data points that are far from other data points. In other words, they're unusual values in a dataset. Outliers are problematic for many statistical analyses because theycan cause tests to either miss significant findings or distort real results.

The standard deviation provides a measure of the overall variation in a data set. To finalize this the relative frequency is needed.

$relativefrequency=\frac{frequency}{totalfrequency}$

The mean of the frequency is given by

$Mean=\frac{18+24+24+22+15+10+5+1}{8}=\frac{119}{8}=14.9$

The standard deviation of the frequency is $8.79$

The standard error of the frequency is $3.1$

So there are only one standard deviation of given data sets can be predicted.

No the answers of part a and part c are different because the median and outliers are calculated in the part a, whereas the mean and standard deviations are calculated from part c.

Histogram helps us to identify the data set shape. the data shape of part c is more accurate than the part a and it has been plotted below.

The shape of data more appropriate for finding the center of the data is median. The median gives the exact data of the center.

From the given data, the median will be calculated for $\text{8}$frequencies.

$Median=\frac{15+18}{2}=16.5$