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3 years ago

Need help with another problem

Mathematics

1 answer:

Verdich [7]3 years ago

4 0

Find the midpoint:

m= x1+x2/2; y1+y2/2

m= 9+-1/2; 8+-2/2

m= 8/2; 6/2

m= (4,3)

(4,3) is your answer.

I hope this helps!
~kaikers

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Answer:

a) $\bar X = 52$

$Median = 55$

b) $Q_1= \frac{44+44}{2}=44$

$Q_3= \frac{57+60}{2}=58.5$

c) $Range = Max -Min = 69-36=33$

d) $s^2 =100.1429$

$s= \sqrt{100.143}=10.0071$

e) $Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25$

$Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75$

A possible outlier would be the value of 69 since its above the upper limti for the boxplot.

Step-by-step explanation:

For this case we have the following dataset:

55 56 44 43 44 56 60 62 57 45 36 38 50 69 65

A total of 15 observations

Part a

We calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^{15} X_i}{15}$

And for this case we got $\bar X = 52$

For the median we ust need to order the data on increasing way like this:

36, 38,43,44,44,45,50,55,56,56,57,60,62,65,69

Since the number of observations is an odd number the median would be on the 8 position from the dataset ordered on this case:

$Median = 55$

Part b

In order to calculate the Q1 we need to select the following data:

36, 38,43,44,44,45,50,55

And the Q1 would be the average between the 4 and 5 positions like this:

$Q_1= \frac{44+44}{2}=44$

And for the Q3 we select these values:

55,56,56,57,60,62,65,69

And the Q3 would be the average between the 4 and 5 positions like this:

$Q_3= \frac{57+60}{2}=58.5$

Part c

The Range is defined as:

$Range = Max -Min = 69-36=33$

Part d

In order to calculate the sample variance we can use the following formula:

$s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

And if we replace we got:

$s^2 =100.1429$

And the deviation is just the square root of the variance:

$s= \sqrt{100.143}=10.0071$

Part e

For this case we need to find the lower and upper limits for the boxplot given by:

$Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25$

$Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75$

A possible outlier would be the value of 69 since its above the upper limti for the boxplot.

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4 years ago