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Nikitich [7]
3 years ago
7

For each part, compare distributions (1) and (2) based on their medians and IQRs. You do not need to calculate these statistics;

simply state how the medians and IQRs compare. Make sure to explain your reasoning
a. (1) 3,5,6,7,9
(2) 3,5,6,7,20

b. (1) 1,2,3,4,5
(2) 6,7,8,9,10

c. (1) 3,5,6,7,9
(2) 3,5,7,8,9

d. (1) 0,10,50,60,100
(2) 0,100,500,600,1000
Mathematics
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

a) The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Step-by-step explanation:

For any distribution,

The median is the variable at the middle when all the variables in the dsitribution are arranged in ascending or descending order.

The median is the (n+1)/2 th variabke.

where n = size of the distribution or number of variables. For these questions, the sample size is 5, hence, the median will always be the 3rd variable when the variables are arranged in ascending or descending order.

The IQR, known as the inter quartile range is a measure of dispersion for the distribution. Although, it isn't as effective as other measures of dispersion such as the standard deviation because the IQR unlike the the standard deviation isn't responsive to changes in the variables, especially ones at the end of the distribution.

The IQR is simply given mathematically as the third quartile minus the first quartile.

IQR = (Third quartile) - (First quartile)

Third quartile is the 3(n+1)/4 th variable. For a sample of n=5, the third quartile is the 4.5th variable, that is the average of the 4th and 5th variable.

First quartile is the (n+1)/4 th variable. For a sample of n=5, the first quartile is the 1.5th variable, that is the average of the 1st and 2nd variable.

Taking the questions one at a time

a) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,6,7,20

Median = 3rd variable = 6

Third quartile = (7+20)/2 = 13.5

First quartile = (3+5)/2 = 4

IQR = 13.5 - 4 = 9.5

The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) (1) 1,2,3,4,5

Median = 3rd variable = 3

Third quartile = (4+5)/2 = 4.5

First quartile = (1+2)/2 = 1.5

IQR = 4.5 - 1.5 = 3

(2) 6,7,8,9,10

Median = 3rd variable = 8

Third quartile = (9+10)/2 = 9.5

First quartile = (6+7)/2 = 6.5

IQR = 9.5 - 6.5 = 3

The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,7,8,9

Median = 3rd variable = 7

Third quartile = (8+9)/2 = 8.5

First quartile = (3+5)/2 = 4

IQR = 8.5 - 4 = 4.5

The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) (1) 0,10,50,60,100

Median = 3rd variable = 50

Third quartile = (60+100)/2 = 80

First quartile = (0+10)/2 = 5

IQR = 80 - 5 = 75

(2) 0,100,500,600,1000

Median = 3rd variable = 500

Third quartile = (600+1000)/2 =800

First quartile = (0+100)/2 = 50

IQR = 800 - 50 = 750

The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Hope this Helps!!!

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If a new data point at 12 is added to the graph , which will be true ?
Orlov [11]

Answer:

3. The mean will increase more than median, but both will increase.

Step-by-step explanation:

We have been given a dot plot. We are asked to find the true statement, if a new data point at 12 is added to our given dot plot.

Since we know that an large valued outlier affects mean more than median, so mean will increase more than median.

Let us check this using our given information. Our data set before adding 12 is:

1, 2, 2, 2, 3, 4, 4, 4, 5.

We can see that our data set has odd number (9) of data points, so median will be the value of 5th term, that is 3.

\text{Mean}=\frac{1+2+2+2+3+4+4+4+5}{9}

\text{Mean}=\frac{27}{9}=3

Therefore, mean of our data set is 3.

Upon adding a data point at 12, our new data set will be:

1, 2, 2, 2, 3, 4, 4, 4, 5, 12.

Now our data set has 10 terms, so median will be the average of 5th and 6th term.

\text{Median}=\frac{3+4}{2}

\text{Median}=\frac{7}{2}=3.5

Therefore, the median of new data set will be 3.5.

\text{Mean}=\frac{1+2+2+2+3+4+4+4+5+12}{10}

\text{Mean}=\frac{39}{10}

\text{Mean}=3.9

Therefore, mean of new data set will be 3.9.

Upon looking at our given statements we can see that 3rd statement is true as mean and median both increased after adding 12 to our data set, but mean increased more that median. Mean and median before adding 12 was 3; but after adding 12 median became 3.5, while mean became 3.9 (which is greater than mean).



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