Statistics | Exercise 13.3

Question 5

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

Find the median life time of a lamp.

Question diagram 1
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Solution
Understand the Question
  • To find the median of grouped data, first construct the cumulative frequency (cfcf) column to determine the total frequency NN.
  • Find N2\dfrac{N}{2} and locate the median class, which is the class interval whose cumulative frequency is just greater than or equal to N2\dfrac{N}{2}.
  • Apply the median formula: Median=l+(N2−cff)×h\text{Median} = l + \left(\dfrac{\dfrac{N}{2} - cf}{f}\right) \times h where ll is the lower limit of the median class, cfcf is the cumulative frequency of the preceding class, ff is the frequency of the median class, and hh is the class size.

Step 1 · Calculate Cumulative Frequency

Diagram 1

Life time (in hours)Number of lamps (f)Cumulative Frequency (cf)1500−200014142000−25005614+56=702500−30006070+60=1303000−350086130+86=2163500−400074216+74=2904000−450062290+62=3524500−500048352+48=400\begin{array}{|c|c|c|} \hline \text{Life time (in hours)} & \text{Number of lamps } (f) & \text{Cumulative Frequency } (cf) \\ \hline 1500 - 2000 & 14 & 14 \\ \hline 2000 - 2500 & 56 & 14 + 56 = 70 \\ \hline 2500 - 3000 & 60 & 70 + 60 = 130 \\ \hline 3000 - 3500 & 86 & 130 + 86 = 216 \\ \hline 3500 - 4000 & 74 & 216 + 74 = 290 \\ \hline 4000 - 4500 & 62 & 290 + 62 = 352 \\ \hline 4500 - 5000 & 48 & 352 + 48 = 400 \\ \hline \end{array}

Total number of lamps: N=400N = 400

Step 2 · Find the Median Class

Calculate N2\dfrac{N}{2}:

N2=4002=200\begin{aligned} \dfrac{N}{2} &= \dfrac{400}{2} \\[0.6em] &= 200 \end{aligned}

The cumulative frequency just greater than 200200 is 216216, which belongs to the class interval 3000−35003000 - 3500.

Median class=3000−3500\text{Median class} = 3000 - 3500

Step 3 · Calculate the Median

From the median class 3000−35003000 - 3500:

  • Lower limit (ll) =3000= 3000
  • Cumulative frequency of preceding class (cfcf) =130= 130
  • Frequency of median class (ff) =86= 86
  • Class size (hh) =3500−3000=500= 3500 - 3000 = 500

Using the median formula:

Median=l+(N2−cff)×h=3000+(200−13086)×500=3000+(7086)×500=3000+3500086=3000+406.9767…=3406.98 hours\begin{aligned} \text{Median} &= l + \left(\dfrac{\dfrac{N}{2} - cf}{f}\right) \times h \\[0.6em] &= 3000 + \left(\dfrac{200 - 130}{86}\right) \times 500 \\[0.6em] &= 3000 + \left(\dfrac{70}{86}\right) \times 500 \\[0.6em] &= 3000 + \dfrac{35000}{86} \\[0.6em] &= 3000 + 406.9767\dots \\[0.6em] &= 3406.98 \text{ hours} \end{aligned}
Answer

3406.98 hours3406.98\text{ hours}

Common Mistakes
  • Using Incorrect cfcf: Mistakenly using the cumulative frequency of the median class (216216) instead of the class preceding the median class (cf=130cf = 130).
  • Confusing ff and cfcf: Mixing up the frequency of the median class (f=86f = 86) with its cumulative frequency.
  • Calculation Error: Errors while dividing 3500086\dfrac{35000}{86}; ensure proper rounding to two decimal places (406.98406.98).

More questions in Exercise 13.3

Q1

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.

Q2

If the median of the distribution given below is 28.5, find the values of xx and yy.

Q3

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.

Q4

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 classes then change to 117.5 - 126.5, 126.5 - 135.5, . . . , 171.5 - 180.5.)

Q5

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

Find the median life time of a lamp.

Q6

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.

Q7

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

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