91Ó°ÊÓ

The given information is:

The values of mean and median for the given data are 34.70 and 27.86 respectively.

We can see that the mean value is higher than the median, from it we can conclude that the skewness of the distribution is towards the right.

Its reason is that the mean may be influenced by the unusual data values if we compare it with the median making the mean value larger than the median.

The standard deviation is a measure of how much the data values deviate from the mean.

The amount spent by 50 grocery shoppers in the sample is expected to vary by 21.70 dollars on average, compared to the mean of 34.70 dollars.

$$\begin{gathered} Q_1=19.27,Q_3=45.40 \\ IQR=45.40-19.27 \\ =26.13 \end{gathered}$$

Observations that are more than 1.5 times the IQR above or below Q1 are considered outliers.

$$\begin{gathered} Q_3+1.5IQR=45.40+1.5\left(26.13\right) \\ =84.595 \\ {Q}_1-1.5IQR=19.27-1.5\left(26.13\right) \\ =-19.925 \end{gathered}$$

We know that the maximum value is 93.34 which is larger than 84.595 making 93.34 an outlier.

Therefore, we can say that outliers are present in this given data.