Fiber content (in grams per serving) and sugar content (in grams per serving) fo

Question

Fiber content (in grams per serving) and sugar content (in grams per serving) for 18 high fiber cereals are shown below Fiber Content 6 10 10 6 9 6 12 12 9 13 10 9 12 6 14 6 9 9 Sugar Content 12 6 14 13 0 18 9 10 19 6 10 17 10 10 0 95 12 (a) Find the median, quartiles, and interquartile range for the fiber content data set. median = lower quartile = upper quartile = interquartile range = (b) Find the median, quartiles, and interquartile range for the sugar content data set. median lower quartile = upper quartile interquartile range = (c) Are there any outliers in the sugar content data set? Yes, there is one outlier on the high end Yes, there are two outliers on the high end Yes, there is one outlier on the low end Yes, there are two outliers on the low end No, there are no outliers (d) Explain why the minimum value for the fiber content data set and the lower quartile for the fiber content data set are equal. The upper quarter of the data values are all the same value. The minimum value and the lower quartile are always the same value The bottom quarter of the data values are all the same value

Explanation / Answer

(a)

Arranging the numbers in ascending order, we get:

6,6,6,6,6,9,9,9,9,9,10,10,10,12,12,12,13,14

n = 18

(i)

Median is average of 9th & 10th items = average of 9 & 9 = 9

So,

Median = 9

(ii) First Quartile:

1/4 (n+1)th item = 1/4 X 19 = 4.75

First Quartile is average of 4th & 5th item = average of 6 & 6 = 6

So,

First Quartile = 6

(iii) Third Quartile:
3/4 (n+1)th item = 3/4 X 19 = 14.25

Third Quarile is average of 14th & 15th item = average of 12 & 12 = 12

Third Quartile = 12

(iv) Inter Quartile Range = Q3 - Q1 = 12 - 6 = 6

(b) Arranging numbers in ascending order, we get:

0,0,5,6,6,9,9,10,10,10,10,12,12,13,14,17,18,19

n = 18

(i) Median:

1/2 (n+2)th item = 1/2 X 19 = average of 9th & 10th item = 10 & 10 = 10

Median = 10

(ii) First Quartile:
1/4 (n+1)th item = 1/4 X 19 = 4.75. Average of 4th & 5th item = 6 & 6 = 6

(iii) Third Quartile

3/4 (n+1)th item = 3/4 X 19 = 14.25 = interpolation of 14th & 15th item = 13 & 14 = 13.25

(iv) Inter Quartile Range = Q3 - Q1 = 13.25 - 6 = 7.25

(c) Lower side outlier:

Q1 - 1.5 IQR = 6 - (1.5 X 7.25) = - 4.875

No value is < - 4.875. So, No outlier on Lower side

Upper side outlier:

Q3+(1.5 IQR) = 13.25 + (1.5 X 7.25) = 24.125

No value is > 24.125.

So,

No outlier on upper side

So,

correct option:

There are no outliers

(d)

Minimum value for fiber content data set = 6

Lower quartile for the fiber content data set = 6

Both are equal = 6

REASON: Correct option: The bottom quarter of the data values are all the same value.

AS PER DIRECTIONS FOR ANSWERING, FIRST 4 QUESTIONS ARE ANSWERED.

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