1. The mean of 20 observations is 50. If one of the observations is 70, find the mean of the remaining 19 observations.
Answer: Let’s solve this step-by-step:
1) Let x be the mean of the remaining 19 observations.
2) We know that the sum of all observations = number of observations × mean
3) For all 20 observations: 20 × 50 = 1000
4) Sum of 19 observations + 70 = 1000
5) 19x + 70 = 1000
6) 19x = 930
7) x = 930 ÷ 19 = 48.95
Therefore, the mean of the remaining 19 observations is 48.95.
2. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting:
a) A king of red colour
b) A face card
c) A red face card
Answer: a) A king of red colour: There are 2 red kings (hearts and diamonds) out of 52 cards. Probability = 2/52 = 1/26 b) A face card: There are 12 face cards (4 each of Jack, Queen, King) out of 52 cards. Probability = 12/52 = 3/13 c) A red face card: There are 6 red face cards (3 each in hearts and diamonds) out of 52 cards. Probability = 6/52 = 3/26
3. The following distribution gives the daily income of 50 workers of a factory:
| Daily income (in ₹) | Number of workers |
|---|---|
| 100 – 120 | 12 |
| 120 – 140 | 14 |
| 140 – 160 | 8 |
| 160 – 180 | 6 |
| 180 – 200 | 10 |
Calculate the median income of the workers.
Answer: Let’s solve this step-by-step:
1) First, we need to find the cumulative frequency:
100 – 120: 12
120 – 140: 12 + 14 = 26
140 – 160: 26 + 8 = 34
160 – 180: 34 + 6 = 40
180 – 200: 40 + 10 = 50
2) The median is the (n+1)/2 th item, where n is the total number of workers.
Here, (50+1)/2 = 25.5 th item
3) The 25.5th item falls in the 120 – 140 class.
4) We use the formula: Median = l + [(n/2 – cf) / f] × h
Where, l = lower limit of median class = 120
n = total frequency = 50
cf = cumulative frequency of class preceding the median class = 12
f = frequency of median class = 14
h = class size = 20
5) Median = 120 + [(50/2 – 12) / 14] × 20
= 120 + [13 / 14] × 20
= 120 + 18.57
= 138.57
Therefore, the median income of the workers is approximately ₹138.57.
4. Two dice are thrown simultaneously. What is the probability of getting:
a) A sum of 8
b) At least one 6
Answer: a) A sum of 8: Favorable outcomes: (2,6), (3,5), (4,4), (5,3), (6,2) Total outcomes: 6 × 6 = 36 Probability = 5/36 b) At least one 6: We can calculate this by subtracting the probability of getting no 6 from 1. Probability of no 6 on either die: (5/6) × (5/6) = 25/36 Probability of at least one 6 = 1 – 25/36 = 11/36
5. The mean of the following frequency distribution is 50. Find the missing frequency f.
| Class | Frequency |
|---|---|
| 0 – 20 | 7 |
| 20 – 40 | f |
| 40 – 60 | 10 |
| 60 – 80 | 9 |
| 80 – 100 | 4 |
Answer: Let’s solve this step-by-step:
1) We use the formula: Mean = Σ(fx) / Σf, where x is the mid-value of each class.
2) Given mean = 50
3) 50 = [7(10) + f(30) + 10(50) + 9(70) + 4(90)] / (7 + f + 10 + 9 + 4)
4) 50(30 + f) = 70 + 30f + 500 + 630 + 360
5) 1500 + 50f = 1560 + 30f
6) 20f = 60
7) f = 3
Therefore, the missing frequency f is 3.