Bihar Board Class 10th Maths Chapter 14 Statistics Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:
Calculation of Mean

We have used direct method because numerical values of xi and fi are small.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory :

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Let the assumed mean, A = 150 and class-size, h = 20.
So, ui = xi−Ah = xi−15020
We construct the table :

E¯ = A + hΣfiuiΣfi = 150 + 20 x −1250
= 150 – 4.8
= 145.2
Hence, mean = Rs 145.20

Question 3.
The following distribution shows the daily pocket allowdnce of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Solution:
Let the assumed mean, A = 16, class size, h = 2,
So, ui = xi−Ah = xi−162

We have : x¯ = 18, A = 16 and h – 2.

Hence, the missing frequency is 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:
Let the assumed mean, A = 75.5 and class size, h = 3.
So, ui = xi−Ah = xi−75.53
We construct the following table :

So, x¯ = A + hΣfiuiΣfi = 75.6 + 3 x 430
= 75.5 + 0.4 = 75.9
Hence, the mean heart beats per minute for these women is 75.9.

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Here, the class intervals are formed by the exclusive method. If we make the data an inclusive one, the mid-values remain same. So, there is no need to convert the data with the caution that while finding h, we should count both the limits of class interval. For example, for 53 – 55, both 53 and 55 should be counted and thus h – 3 and not 55 – 53 = 2.
Let the assumed mean be A = 60 and h = 3.
So, ui = xi−Ah = xi−603
Calculation of Mean

∴ x¯ = A + hΣfiuiΣfi = 60 + 3 x −375400
= 60 – 2.8125 = 57.1875 = 57.19 (nearly)
Hence, mean number of mangoes per box is 57.19.
Here, we have used the step-deviation method for finding the mean.

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.
Solution:
Let the assumed mean, A = 225, class size, h = 50
So, ui = xi−Ah = xi−22550
We construct the following table :

So, x¯ = A + hΣfiuiΣfi
= 225 + 50 x −725
= 225 – 14 = 211
Hence, the mean daily expenditure on food is Rs 211.

Question 7.
To find the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below :

Find the mean concentration of SO2 in the air.
Solution:
Calculation of mean by direct method

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:
Here, the class size varies, and xi’s are small.
Let us apply the direct method here :

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:
Let the assumed mean, A = 70 and class size, h = 10.
So, ui = xi−Ah = xi−7010

Thus, mean literacy rate is 69.43%.

BSEB Bihar Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year :

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
The class 35 – 45 has the maximum frequency. Therefore, this is the modal class.
Here l = 35, h = 10, f1 = 23, f0 = 21 and f2 = 14.

Calculation of Mean
Let assumed mean, A = 30 and class size, h = 10.

Hence, Mode = 36.8 years
and Mean = 35.37 years.
Maximum number of patients admitted in the hospital are of age group 36.8 years (approx,), while on an average the age of a patient admitted to the hospital is 35.37 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.
Solution:
The class 60 – 80 has the maximum frequency. Therefore, this is the modal class.
Here, l = 60, h = 20, f1 = 61, f0 = 52 and f2 = 38.
Now, let us substitute these values in the formula

Thus, the modal lifetimes of the components is 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Solution:
The class 1500 – 2000 has the maximum frequency. Therefore, this the modal class.
Here, l = 1500, h = 500, f1 = 40, f0 = 24 and f2 = 33.
Now, let us substitute these values in the formula

∴ Modal monthly expenditure = Rs 1847.83
Let the assumed mean be A = 3250 and h = 500.

Hence, the mean expenditure is Rs 2662.50.

Question 4.
The following monthly distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:
The class 30 – 35 has the maximum frequency. Therefore, this is the modal class.
Here, l = 30, h = 5, f1 = 10, f0 = 9 and f2 = 3.
Now, let us substitute these values in the formula

Calculation of mean:

Thus, most states/U.T. have a student teacher ratio of 30.6 and on an average, the ratio is 29.2.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches :

Find the mode of the data.
Solution:
The class 4000 – 5000 has the maximum frequency. Therefore, this is the modal class.
Here, l = 4000, h = 1000, f1 = 18, f0 = 4 and f2 = 9.
Now, let us substitute these values in the formula

Thus, the mode of the given data is 4608.7 runs.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Solution:
The class 40 – 50 has the maximum frequency. Therefore, this is the modal class.
Here, l = 40, h = 10, f1 = 20, f0 = 12 and f2 = 11.
Now, let us substitute these values in the formula

BSEB Bihar Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Calculation of median
First, we prepare the table given below to compute the median:

We have : n = 68 Now, n2 = 682 = 34
The cumulative frequency, just greater than n2, is 42 and the corresponding class is 125 – 145. Thus, 125 – 145 is the median class such that
n2 = 34, l = 125, c.f. = 22, f= 20 and h = 20.
Substituting these values in the formula

Calculation of mean
Let the assumed mean, A = 135 and class size, h = 20.

Calculation of mode:
The class 125-145 has the maximum frequency. Therefore this is the modal class.
Here, l = 125, h = 20, f1 = 20 f0 = 13 and f2 = 14.
Now, let us substitute these values in the formula

Clearly, the three measures are approximately the same in this case.

Question 2.
If the median of the distribution given below is 28.5, find the values pf x and y.

Solution:
Here, it is given that median is 28.5.
and n = ∑fi = 60
We now prepare the following cumulative frequency table :

Here n = 60. So, n2 = 30
Since the median is given to be 28.5, thus the median class is 20 – 30.
∴ l = 20, h = 10, f= 20 and cf = 5 + x
∴ Median = l + (n2−cff) x h
or 28.5 = 20 + 30−(5+x)20 x 10
or 28.5 = 20 + 25−x2
or 57 = 40 + 25 – x
or x – 65 – 57 = 8
Also, 45 + x + y = 60
So, 45 + 8+y = 60 [∵ x = 8]
or y = 60 – 53 = 7
Hence, x = 8 and y = 7.

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Solution:
We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute median.

Here, n = ∑fi = 100 So, n2 = 50.
We see that the cumulative frequency just greater than n2, i.e., 50 is 78 and the corresponding class is 35 35 – 40 is the median class.
∴ n2 = 50, l = 35, cf = 45, f = 33 and h = 5.
Now, let us substitute these values in the formula

Hence, the median age is 35.76 years.

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Find the median length of the leaves.
Solution:
Here the frequency table is given in the inclusive form. So, we first convert it into exclusive form by subtracting and adding n2 to the lower and upper limits respectively of each class, where h denotes the difference of lower limit of a class and the upper limit of the previous class. Converting the given table into exclusive form and preparing the cumulative frequency table, we get

We have : n = 40 So, n2 = 20.
The cumulative frequency just greater than n2 is 29 and the corresponding class is 144.5 – 153.5.
So, 144.5 – 153.5 is the median class.
Here n2 = 20, l = 144.5, h = 9, f= 12 and cf = 17.
Substituting these values in the formula

Question 5.
The following table gives the distribution of the life time of 400 neon lamps :

Find the median life time of a lamp.
Solution:
First, we prepare the following table to compute the median :

We have : n = 400 So, n2 = 200.
The cumulative frequency just greater than n2 is 216 and the corresponding class is 3000 – 3500. Thus, 3000 – 3500 is the median class such that n2 = 200, l = 3000, cf = 130, f = 86 and h = 500.
Substituting these values in the formula

Hence, median life time = 3406.98 hours

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows :

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
[Annual Paper {Delhi) 2008]
Solution:
Calculation of median
First, we prepare the following table to compute the median :

We have : n = 100. So, n2 = 50.
The cumulative frequency just greater than n2 is 76 and the corresponding class is 7 – 10. Thus, 7 – 10 is the median class such that
n2 = 50, l = 7, cf = 36, f = 40 and h = 3.
Substituting these values in the formula

Calculation of mean

Calculation of mode.
The class 7 – 10 has the maximum frequency. Therefore this is the modal class.
Here, 1 = 1, h = 3, f1 = 40, f0 = 30 and f2 = 16.
Now, let us substitute these values in the formula

Hence, median = 8.05, mean = 8.32 and mode = 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution:
Let us prepare the following table to compute the median :

We have : n = 30. So, n2 = 15
The cumulative frequency just greater than n2 is 19 and the corresponding class is 55 – 60.
Thus, 55 – 60 is the median class such that n2 = 15, l = 55, f = 6, cf = 13 and h = 5.
Substituting these values in the formula

Hence, the median weight is 56.67 kg.

BSEB Bihar Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory :

Convert the distribution above to a less than type, cumulative frequency distribution and draw its ogive.
Solution:
Converting the given distribution to a less than type cumulative frequency distribution, we get:

Let us now plot the points corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on a graph.paper and join them by a free hand smooth curve.

The curve thus obtained is the required less than ogive.

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows :

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
Here the values 38, 40, 42, 44, 46, 48, 50 and 52 are the upper limits of the respective class-intervals. To represent the data in the table graphically, we mark the upper limits on the class intervals on x-axis and their corresponding cumulative frequencies on the y-axis, choosing a convenient scale.

Let us now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them by a free hand smooth curve.
The curve thus obtained is the less than type ogive
To locate the median point on the graph, take n2 = 352 = 17.5 on the y-axis. From this point draw a line parallel to the x-axis cutting the curve at a. point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis gives the median of the data. In this case it is 46.5.

Let us make the following table in order to find median by using the formula :

Here, 352 , i.e., 17.5 will be in 46 – 48. Therefore 46 – 48 is the median class.
Here, l = 46, n2 = 352, cf = 14, f = 14
So, using the formula, median = l + (n2−cff) x h
Median = 46 + (17.5−1414 x 2
= 46.5
Thus, median is verified.

Question 3.
The following tables gives production yield per hectare of wheat of 100 farms of a village :

Change the distribution to a more than type distribution, and draw its ogive.
Solution:
Converting the given distribution to a more than type distribution, we get

Now, draw the ogive by plotting the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on the graph paper and join them by a free hand smooth curve.

The curve thus obtained is the required more than type ogive.