Q) BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said ‘BINGO’.
Based on the above information, answer the following:
(i) Write the median class.
(ii) When first ball was picked up, what was the probability of calling out an even number?
(iii)(a) Find median of the given data.
      OR
      (b) Find mode of the given data.

Ans:

(i) Median class:

Let’s re-organize the data in the frequency table to find out each part:

To find the median, we need to identify middle value of the data. Let’s rearrange the data:

• First, we need to find the cumulative frequency in the frequency table to find the median. Its shown in last column.
• Next, Total number of sub-divisions or Sum of the frequencies = 48. It shown in the last row of middle column.
• Next, we need to identify Median Class. Since the Median class is the class where the cumulative frequency crosses 50% of the half the total number of sub-divisions. Here in the table, Cumulative frequency of 27 is crossing 50% of frequency i.e. 24, at class “30 – 45”.

Hence, our Median class = 30 – 45

(ii) Probability of even number:

There are total 75 balls with 75 numbers written. Hence, our total number of chances are 75.

We need to draw out an even number. Since there are 37 even numbers from 1 to 75, hence our favourable chances are 37.

We know that the probability =

By putting the values, we get probability =

Hence, our probability of drawing out an even number is .

(iii) (a) Median of the data:

To find the median, we use the formula:

Median = L+x h

Here:

L = Lower boundary of the median class = 30

n = Total number of sub-divisions = 48

= Cumulative frequency of the class before the median class = 17

f = Frequency of the median class = 10

h = Class width = 45 – 30 = 15

hence, the Median = 30 + x 15

⇒ 30 + [(24 – 17)] x

⇒ 30 +

⇒ 30 + 10.5 = 40.5

Therefore, Median of the given data is 40.5.

(iii)(b). Mode of the data:

Since the modal class is the class with the highest frequency.

In the given question, class “45 – 60” has 12 frequency which is the highest frequency among all other classes. Hence, modal class is “45 – 60”.

Now mode of the grouped data is calculated by:

Mode = L + x h

Here,

L = lower class limit of modal class = 45

 = frequency of modal class = 12

 = frequency of class proceeding to modal class = 10

 = frequency of class succeeding to modal class = 9

h = class size = 60 – 45 = 15

Let’s put values in the formula and solve:

Mode = L + x h

= 40 + x 15

= 40 + x 15

= 40 + 6  = 46

Hence, the mode value is 46.

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