5 Draw a histogram of ++
(a)
horam of the distribution and estimate its mode from the histogram.
Weight (in kg) 50–55 55–60 60-65 65-70 70-75 75-80
No. of people 10 28 40 30 18 8
Answers
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 years.
Sol. The given table is cumulative frequency distribution. We write the frequency distribution as given below:
We have
The cumulative frequency just greater than i.e., just greater than 50 is 78.
The median class is 78.
Now,
Now, Median =
Median = 35 +
Thus, the median age = 35.76 years.
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.
[Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The class then change to 117.5–126.5, 126.5–135.5, ..., 171.5–180.5.]
Sol. After changing the given table as continuous classes we prepare the cumulative frequency table:
Now,
The cumulative frequency just above i.e., 20 is 29 and it corresponds to the class 144.5–153.5.
So, 144.5–153.5 is the median class.
We have:
and h = 9
Median =
= 146.75 mm.
5. The following table gives the distribution of the life time of 400 neon lamps:
Find the median life time of a lamp.
Sol. To compute the median, let us write the cumulative frequency distribution as given below:
We have
Since, the cumulative frequency just greater than i.e., greater than 200 is 216.
The median class is 3000–3500
l = 3000, cf = 130, f = 86, h = 500
and
Now, Median =
= 3406.98
Thus, median life = 3406.98 hours.
6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the 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.
Sol. Median
Since, the cumulative frequency just greater than i.e., greater than 50 is 76.
The class 7 – 10 is the median class,
We have
l = 7
cf = 36
f = 40
and h = 3
Median =
= 7 + 1.05 = 8.05
Mean:
Mode
Since the class 7 – 10 has the maximum frequency.
The modal class is 7 – 10
So, we have
l = 7, h = 3
f1 = 40
f0 = 30
f2 = 16
Mode =
= 7.88
Thus, the required
Median = 8.05, Mean = 8.32 and Mode = 7.88.
7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Sol. We have
n = 30
The cumulative frequency just more than i.e., more than 15 is 19, which corresponds to the class 55–60.
l = 15
f = 6
cf = 13 and h = 5
Median =
Thus, the required median weight = 56.67 kg.
EXERCISE : 14.4
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.
Sol. We have the cumulative frequency distribution as:
Now, we 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 as shown below:
The curve so obtained is called the less than ogive.
2. During the medical check-up of 35 students of a class, their wrights 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.
Sol. Here, the values 38, 40, 42, 44, 46, 48, 50 and 52 are the upper limits of the respective class-intervals.
We plot the points (ordered pairs) (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 so obtained is the less than type ogive.
From the point (i.e., from 17.5) we draw a line parallel to the x-axis which cuts the curve at P. From this point (i.e., from P), draw a perpendicular to the x-axis, meeting the x-axis at Q. The point Q represents the median of the data which is 47.5.
Verfication
To verify the result using the formula, let us make the following table in order to find median using the formula:
Here,
Since, the observation lies in the class 46 – 48.
The median class is 46 – 48 such that
l = 46, h = 2, f = 14, cf = 14
Median =
= 46.5
Thus, the median = 46.5 kg is approximately verified.
3. The following table 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.
Sol. For `more than type' distribution, we have:
Now, we plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) and join the point with a free hand curve.
The curve so obtained is the `more than type ogive'.