Statistics Test - 16

In a given class 90-120, upper class = 120 and lower class = 90

We know that, class mark = \({Upper\, class\, + Lower\, class\,}\over 2\)

= \({120 +90}\over 2\)

= \(\frac{210}{2}\)

= 105

Maximum value of the variate = 32

And the minimum value of the variate = 6

Range = Maximum value of the variate – Minimum value of the variate

= 32 – 6

= 26

Let x and y be the upper and lower class limit in a frequency distribution.

Now, mid value of a class  (x + y )/2 =10  [given]

x + y = 20 …(i)

Also, given that, width of class  x- y = 6 …(ii)

On adding Eqs. (i) and (ii), we get

2x = 20+ 6

2x = 26

x = 13

On putting x = 13 in Eq. (i), we get

13 + y = 20

y = 20 - 13

y = 7

Hence, the lower limit of the class is 7.

Width of each of the five continuous classes in a frequency distribution is 5.

Lower class limit of the lowest class = 10

Upper class limit of the lowest class is 10 + 5 = 15

So, the five continuous classes are

10 – 15, 15 – 20, 20 – 25, 25 – 30, 30 – 35

Hence, the upper-class limit of the height class is 35.

The class mark are 15, 20, 25, …. The size of each class interval is 25 – 20 = 20 – 15 = 5

Hence, the class interval corresponding to the class mark 20 is

(20 – 2.5) – (20 + 2.5) i.e., 17.5 – 22.5

Since, the class interval 10-20 is the first interval of frequency distribution and 20-30 is the next one but the number 20 is present in both intervals. We know that, the presence of 20 in the interval 10-20 is not fully 100% while in the next interval 20-30, presence of it fully 100%.

The observation corresponding to class 310 – 330 (330 not included in this interval) are 310, 310, 320, 319, 318, 316, i.e., 6 observations.

Hence, the frequency of the class 310 – 330 is 6.

Mean of five numbers is 30.

Sum of five number = 30 x 5 = 150

Let the excluded number be x .

Then, according to question,

Mean of four numbers = \({150 -x}\over 4\) = 28

150 - x = 112

x = 150 -112 = 38

If each observation of the data is increased by 5, then their mean is also increased by 5.

Mean of first 13 observations = 32

Sum of all first 13 observations = (32 x 13) = 416

Mean of last 13 observation = 40

Sum of all last 13 observations = (40 x 13) = 520

Mean of 25 observations = 36

Sum of all first 25 observation = (36 x 25) = 900

Hence, 13th observation = 416 + 520 - 900 = 36

Arranging the data in ascending order, we get

22, 34, 39, 45, 54, 54, 56, 78, 84

Here n = 9, which is an odd number.

Median = \(({{n + 1}\over 2})th\)

Value = \(({{9 + 1}\over 2})th\) 

Value = 5th value

So, median = 54

Class marks i.e., the mid-point of the classes are abscissa of the points, which we plot for frequency polygon.

We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20 From above, we see that 15 occurs most frequently i.e., 5 times.

Hence, the mode of the given data is 15.

Arrange the data m ascending order 32, 33, 36, 42, 48, 50, 52, 53, 56, 60, 70, 75, 80, 95.

Here, total number of observations is fifteen which is odd. Therefore, media \(({ {n + 1} \over 2})th\) 

Observation = \(n({{15+1}\over2})th\)  observation = 8th observation that is 52.

In a frequency distribution, the mid value of a class is 15 the width of the class is 4. The lower limit of the class is 13.

The first eight natural numbers are 1, 2, 3, 4, 5, 6, 7 and 8.

Mean = \(\frac{Sum\, of\, observations\,}{Number\, of\, observations\,}\)

= \({1 +2 + 3 +4 + 5 + 6 + 7 + 8} \over 8\)

= \(\frac{36}{8}\)

= 4.5

Large data is condensed called is Classes.

First ten prime numbers in ascending order are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29

Here n =10, which is even

Median = \(({n\over2})^{th}\, item + ({{n\over2} + 1})^{th}\, item \over 2\)

= \(({5})^{th}\, item + ({6})^{th}\, item \over 2\)

= \({11 +13} \over 2\)

= \(\frac{24}{2}\)

= 12

No. of boys in a school = 90

Mean marks of boys = 45

Total mark of 90 boys = 90 x 45 = 4050

No. of girls in a girls =70

Total marks of 30 girls

Total = 30 x 70 = 2100

Total marks secured by 90 boys and 30 Girls = 4050 + 2100 = 6150

No. of boys and girls in a school = 90 + 30 = 120

Average marks % of school = \(\frac{6150}{120}\) = 51.25

The mid-point of the top sides of rectangles represent the mid values of classes for which the histogram is drawn. So, the required graph is a frequency polygon.