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postnew [5]
4 years ago
5

What is the range of the data in this stem-and-leaf plot? 2 14 19 32.5 Stem and leaf plot. Vertical line separates each stem fro

m its first leaf. First row. Stem 2. Leaves 6, 7, 7. Second row. Stem 3. Leaves 0, 2, 3, 3, 4. Third row. Stem 4. Leaves 0, 5. Key. 3 vertical line 0 equals 30.
Mathematics
2 answers:
inysia [295]4 years ago
4 0

Answer:

19

Step-by-step explanation:

Given :

Stem   Leaves

2    |      6, 7, 7

3    |      0, 2, 3, 3, 4.

4    |      0, 5

To Find: What is the range of the data in this stem-and-leaf plot?

Solution :

Since we are given key 3 vertical line 0 equals 30.

So, in the given stem leaf plot the number are :

26,27,27,30,32,33,33,34,40,45

Now Range = maximum number   -  minimum number

Range = 45-26

Range = 19

Hence The range of the data in this stem-and-leaf plot is 19

dsp734 years ago
3 0

Answer:

amos grant.e97

Step-by-step explanation:

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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.

4 0
4 years ago
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