Chapter SH, Problem 13.1MPS

i.

To determine

Calculate the mean, median and mode for given classes.

  

Answer to Problem 13.1MPS

For first period,

Mean =83.7

Median = 85

Mode = 90

For second period,

Mean =82.3

Median = 81

Mode = 92

Explanation of Solution

Given:

  

Calculations:

Here, we have to calculate mean, median and mode for given data.

Mean is the average of given set of data.

Mean for first period is $\frac{\left(92+72+98+90+96+90+84+98+90+74+82+72+64+86+76+88+70+90+80+82\right)}{20}=83.7$

Mean for second period is $\frac{\left(90+70+98+60+76+80+74+92+92+94+80+78+82+80+84+90+92+78+74+82\right)}{20}=82.3$

Median is the centered value of given data.

Median for first period is

64, 70, 72, 72, 74, 76, 80, 82, 82, 84, 86, 88, 90, 90, 90, 90, 92, 96, 98, 98

  $\frac{84+86}{2}=85$

Median is the centered value of given data.

Median for first period is

60, 70, 74, 74, 76, 78, 78, 80, 80, 80, 82, 82, 84, 90, 90, 92, 92, 92, 94, 98

  $\frac{80+82}{2}=81$

Mode is the frequent value in given data.

Thus, in first period, 90 is mode.

In second period, 92 is mode.

Conclusion:

Therefore, we are able to find mean, median and mode of given data.

ii.

To determine

Draw stem and leaf plot for given data.

  

Answer to Problem 13.1MPS

Back to back stem and leaf plot for given data is

First period	Stem	Second period
4	6	0
6 4 2 2 0	7	0 4 4 6 8 8
8 6 4 2 2 0	8	0 0 0 2 2 4
8 8 6 2 0 0 0 0	9	0 0 2 2 2 4 8

Explanation of Solution

Given:

  

Calculations:

Stem and leaf plots shows the frequency of given data.

There are two parts of this plot namely stem and leaf. Stem shows first digit and leaf shows last digit.

We have given two different data. For this we has to draw back to back stem and leaf plot.

First period	Stem	Second period
4	6	0
6 4 2 2 0	7	0 4 4 6 8 8
8 6 4 2 2 0	8	0 0 0 2 2 4
8 8 6 2 0 0 0 0	9	0 0 2 2 2 4 8

Conclusion:

Therefore, we are able to draw back to back stem and leaf plot.

iii.

To determine

Calculate measures of variation for given data.

  

Answer to Problem 13.1MPS

Measures of variation for given data is,

Measures of variation	First period	Second period
Range	34	38
Median	75	81
Upper Quartile	75	77
Lower Quartile	90	91
Interquartile range	15	14

Explanation of Solution

Given:

  

Calculations:

Here, we have to calculate measures of variation and outlier for given set of data.

Measures of variation include range, upper quartile, lower quartile and interquartile range of given data.

Range is the difference between higher and lower data value.

Here, (Range) _Firstperiod = 98-64 = 34

(Range)_Secondperiod = 98-60 = 38

Median for first period is 85 and for second order is 81.

Upper quartile is the median of upper half set of data.

Here, (Upper quartile)_Firstperiod is, $\frac{74+76}{2}=75$

(Upper quartile)_Secondperiod is, $\frac{76+78}{2}=77$

Lower quartile is the median of lower half set of data.

Here, (lower quartile) _Firstperiod is, $\frac{90+90}{2}=90$

(Lower quartile)_Secondperiod is, $\frac{90+92}{2}=91$

Interquartile range is the difference between upper quartile and lower quartile.

Here, (Interquartile range) _Firstperiod =90-75=15

(Interquartile range)_Secondperiod =91-77=14

Conclusion:

Therefore, we are able to find measures of variation for given data.