Question : 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 participants 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'
| Numbers Announced | Number of times |
| 0-15 | 8 |
| 15-30 | 9 |
| 30-45 | 10 |
| 45-60 | 12 |
| 60-75 | 9 |
Based on the above information, answer the following:
(i) Write the median class. [1 Mark]
(ii) When first ball was picked up, what was the probability of calling out an even number? [1 Mark]
(iii) (a) Find median of the given data. [2 Marks]
OR
(iv) Find mode of the given data. [2 Marks]
CBSE Delhi 2024 (30/1/1)
Solution :
This is a case study question based on Statistics and Probability.
(i)
| Numbers Announced | Number of times | Cumulative Frequency (CF) |
| 0-15 | 8 | 8 |
| 15-30 | 9 | 17 |
| 30-45 | 10 | 27 |
| 45-60 | 12 | 39 |
| 60-75 | 9 | 48 |
| ∑fi=48 |
Here N=∑fi=48
N/2=48/2=24
N/2=48/2=24
Median Class = 30-45
(ii) Total Number of Balls {1,2,3,. . . , 75}
= 75-1+1 = 75
Number of balls having even number {2,4,6. . .74}
=37
= 75-1+1 = 75
Number of balls having even number {2,4,6. . .74}
=37
P(E) = [Number of favorable outcomes] / [Total number of possible outcomes]
P(Calling out an even number)
= [No. of balls having even number]/[Total number of balls]
= 37/75
= [No. of balls having even number]/[Total number of balls]
= 37/75
(iii)
(a) Median
| Numbers Announced | Number of times | Cumulative Frequency (CF) |
| 0-15 | 8 | 8 |
| 15-30 | 9 | 17 |
| 30-45 | 10 | 27 |
| 45-60 | 12 | 39 |
| 60-75 | 9 | 48 |
| ∑fi=48 |
Here N=∑fi=48
N/2=48/2=24
N/2=48/2=24
Median Class = 30-45
Lower Limit of the Median Class (L) = 30
Class Size (h)
=45-30
=15
CF=17
f=10
Class Size (h)
=45-30
=15
CF=17
f=10
Median
= L+[(N/2-CF)/f]×h
= 30+[(24-17)/10]×15
= 30+[7/10]×15
= 30+[0.7×15]
= 30+10.5
= 40.5
= L+[(N/2-CF)/f]×h
= 30+[(24-17)/10]×15
= 30+[7/10]×15
= 30+[0.7×15]
= 30+10.5
= 40.5
Hence, the median of the given data is 40.5
OR
(b)
Mode
| Numbers Announced | Number of times |
| 0-15 | 8 |
| 15-30 | 9 |
| 30-45 | 10=f0 |
| 45-60 | 12=f1 |
| 60-75 | 9=f2 |
Here, the maximum frequency is 12
Modal Class = 45-60
Modal Class = 45-60
Lower limit of the modal class (L) = 45
Class size (h) = 60-45 = 15
f0=10
f1=12
f2=9
Mode
= L+[(f1-f0)/(2f1-f0-f2)]×h
= 45+[(12-10)/(2{12}-10-9)]×15
= 45+[(2)/(24-19)]×15
= 45+[2/5]×15
= 45+[2]×3
= 45+6
= 51
Hence, the mode of the given data is 51.
= L+[(f1-f0)/(2f1-f0-f2)]×h
= 45+[(12-10)/(2{12}-10-9)]×15
= 45+[(2)/(24-19)]×15
= 45+[2/5]×15
= 45+[2]×3
= 45+6
= 51
Hence, the mode of the given data is 51.