Question

on se All Questions From Part 0 A. B. Given an example of a sm all set of data for which theme an is larger than third quartile The table provides the ratio of omega-3 to omega-6 fatty acids in common food oils. The distribution is strongly right-skewed with a high outlier Omega-3 fatty acids as a fraction of omega-6 fatty acids in food oils Oil Perilla Walnut Wheat germ Mustard Sardine Salmon Mayonnaise Cod liver Ratio 5.33 0.20 0.13 Oil Margarine Olive Shea nut Ratio 0.05 0.08 Soybean, hydrogenated 0.07 0.06 Oil Ratio Herring 2.67 0.05 0.64 0.03 0.01 0.01 0.00 0.02 0.04 Rice bran Butter Sunflower Corn Sunfower (oleic 0.05 2.50 0.06 2.00 Shortening (household) 0.11 Shortening (industrial) 0.06 2.16 Sunflower lnoleic) 0.00 Flaxseed Canola Soybean Grape seed 3.56 0.46 0.13 0.00 1.96 Palm Cocoa butter Calculate the mean. (Round your answ er to four decimal places.) Calculate the median. (Round your answ er to three decimal places.) Choose the best explanation for the phenomenon you observed. The mean and the median are about the same because the distribution is strongly right-skewed with a high outlier The mean is larger than the median because the distribution is strongly right-skewed with a high outlier The distribution is strongly right-skewed with a high outlier but that does not influence whether the medi an is larger than the mean or not The median is larger than the mean because the distribution is strongly right-skewed with a high outlier

Answer 1

Solution:

A. Given an example of a small set of data for which the mean is larger than the third quartile.

Answer: Suppose we have the following data about the weight of a group of 7 persons. The data is:

$\small 62,65,68,69,72,71,104$

The mean of given data is:

$\small Mean=\frac{62+65+68+69+71+72+104}{7}=73$

$\small Q_{3}=\left ( \frac{3(N+1)}{4} \right )^{th }item=\left (\frac{3 \times (7+1)}{4} \right )^{th} item=6^{th }item=72$

Therefore, the mean is larger than the third quartile.

B.

Calculate the mean.

Answer:

The mean is given below:

$\small Mean =\frac{\sum x}{n}$

  $\small =\frac{22.82}{30}$

$\small =0.7607$

The median is given below:

First, we need to sort the data in ascending order of magnitude as:

$\small 0,0,0,0.01,0.01,0.02,0.03,0.04,0.05,0.05,0.05,0.06,0.06,0.06,0.07,0.08,0.11,0.13,0.13,0.2,0.38,0.46,0.64,1.96,2,2.16,2.5,2.67,3.56,5.33$The median is given below:

$\inline \small Median = \left ( \frac{N+1}{2} \right )^{th} item= \left ( \frac{30+1}{2} \right )^{th} item=15.5^{th} item$

  $\inline \small =15^{th}item +0.5(16^{th}item-15^{th}item)$

$\inline \small =0.07 +0.5(0.08-0.07)$

$\inline \small =0.075$

Therefore, the median is $\inline \small \boldsymbol{0.075}$ .

Choose the best explanation for the phenomenon you observed.

Answer: The mean is higher than the median because the distribution is strongly right-skewed with a high outlier.