Chapter- Probablity,Statistics
Ques:1) Mean of 12 observation is 17. If each observation is decreased by 2. Find the new mean.(Mark-1)
Solution:
For mean (x̅),
x̅ = (Sum of all obs.)/(Total no. of obs.)
17 = (Sum of all obs.)/12
Sum of all obs. = 17 × 12
= 204
Now, When each obs. is decreases by 2,
Then,
New sum of obs. =204 – Total obs. × 2
= 204 – 12×2
= 204-24
New sum of obs.= 180
Total no. Observations = 12
Therefore, New mean (x̅)
x̅ = (Sum of all obs.)/(Total no. of obs.)
x̅ = 180/12
x̅ =15
Ques:2) Find the class size of the class – intervals 2.5-5.5, 5.5-8.5,….(Mark-1)
Solution:
In 1st Class interval,(2.5-5.5)
Class size = Upper Value – lower value
= 5.5 – 2.5
Class size = 3.5
Similarly, In 2nd Class interval,
Class size = Upper Value – lower value
= 8.5 – 5.5
Class size = 3.5
class size of the class – intervals 2.5-5.5, 5.5-8.5,… is 3.5
Ques:3)A coin is tossed 1000 times with the following frequencies.
Head : 450, Tail : 545
Find the probablity of getting tail.(Marks-2)
Solution:
Total no.Of times Coin tossed = 1000
Frequency of getting head = 450
Frequency of getting tail = 545
Probablity of getting tail
=(Frequency of tail)/(Total no.Coin tossed)
= 545/1000
= 0.545
Therefore,
Probablity of getting tail is 0.545
Ques:4)Find the probablity of getting an ace when one card is drawn from a well shuffled deck of 52 cards.(Marks-2)
Solution:
Total no. Of cards = 52
Frequency of getting an ace in 52 deck of cards = 4
Probablity of getting an ace,
=(Freq. of getting an Ace)/(Total no.Cards)
= 4/52
= 1/13
Therefore,
Probablity of getting an ace is 1/13.
Ques:4)The following number of goals were scored by a team in a series of 10 matches.
2,3,4,5,0,1,3,3,4,3
Find the mean,median and mode of these scores.(Marks-3)
Solution:
Sum of all observations = 28
Total no. Of observations (n)= 10
For mean (x̅),
x̅ = (Sum of all obs.)/(Total no. of obs.)
x̅ = (2+3+4+5+0+1+3+3+4+3)/10
x̅ = 28/10
x̅ = 2.8
Mean score = 2.8
Now,
Arrange the scores in ascending order, 0,1,2,3,3,3,3,4,4,5
Since, n is even,
Therefore, median will be mean of n/2 and (n/2)+1
Or, n/2 = 10/2
= 5th observation
Also, (n/2)+1 = 10/2 + 1
= 6th observation
For Median,
Median = (5th obs.+ 6th obs.)/2
= (3+3)/2
= 6/2
Median score = 3
Mode of data is the observation with the maximum frequency in data.
Therefore, Mode of the score is 3 , as it has the maximum frequency in data.
Ques:7) Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops in noted and recorded in the table given below.
Sum. Frequency.
2 14
3. 30
4. 42
5. 55
6. 72
7. 75
8. 70
9. 53
10. 46
11. 28
12. 15
What is the probablity of getting a sum
I) 3?
II) more than 10?
III) less than or equal to 5?
IV) between 8 and 12? (Marks-4)
Solution:
Total no. of dices are thrown = 500
(i) Frequency of getting sum 3 on two dices = 30
Probablity of getting sum 3 =
=(frequency of getting sum 3)/(Total no. Of dices thrown)
= 30/500
= 0.06
Therefore, Probablity of getting sum 3 is 0.06.
(ii)Frequency of getting sum more than 10 = 46+28+15 = 89
Probablity of getting sum more than 10 =(frequency of getting sum more than 10)/(Total no. Of dices thrown)
=89/500
=0.178
Therefore,Probablity of getting sum more than 10 is 0.178
(iii)Frequency of getting sum less than or equal to 5 = 55+42+30+14 = 141
Probablity of getting sum less than or equal to 5
= (frequency of getting sum less than or equal to 5)/(Total no. Of dices thrown)
= 141/500
= 0.282
Therefore,Probablity of getting sum less than or equal to 5 is 0.282
(iv) Frequency of getting sum b/w 8 and 12 = 53+46+28 = 127
Probablity of getting sum b/w 8 & 12=
= (frequency of getting sum b/w 8 & 12)/(Total no. Of dices thrown)
= 127/500
= 0.254
Therefore, Probablity of getting sum b/w 8 & 12 is 0.254
Ques:8) The blood groups of 35 students of class IX were surveyed and recorded as below:
Blood group No. Of Students
A. 11
B. 8
AB. 7
O. 9
Total. 35
If a student is chosen at random, find the probablity that he/she has blood group B or A.(Marks-2)
Solution:
Total no. Of students in class = 35
Frequency of students having blood group B or A = 11 + 8 = 19
Probablity of students having blood group B or A =
= (Frequency of students having blood group B or A)/(Total no. Of students in class)
= 19/35
= 0.542
Therefore,Probablity of students having blood group B or A is 0.542
Ques:9) Eleven bags of wheat flour contains the following weights of flour(in kg): 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 5.07, 5.04, 5.00, 4.98. Find the probablity that any of these bags chosen at random contains more than 5 kg of flour.(Marks-3)
Solution:
Total no. Of bags of Wheat flour = 11
No. of bags containing more than 5 kg of wheat flour = 7
Probablity of bag containg more than 5 kg of wheat flour =
=(No. of bags containing more than 5 kg of wheat flour)/(Total no. Of bags)
= 7/11
= 0.636
Therefore,Probablity of bag containg more than 5 kg of wheat flour is 0.636
Apex Class 
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