Download Grade 7 Data Handling - Probability, Statistics

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Grade 7
Data Handling - Probability, Statistics
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Answer t he quest ions
(1)
Following f igure shows f ruits in a basket at Samira's house. Samira is getting late f or school,
and just picks a f ruit f rom the basket without looking at it. Which f ruit is most likely to be picked
?
(2)
(3)
From a deck of cards, Sadia withdraw a card at random. What is the probability that the number
on the card is a prime number?
If there are n numbers of which one is
and all the others are 1's, then by how much
is the arithmetic mean of these numbers less than 1.
(4) If 80C - 240A = 950, and 320A + 80B = 1690, then what is the average of A, B and C?
(5)
Ayaaz tosses a die and tosses a coin. What is the probability that he gets a 4 on the die and a
T ail f rom the coin toss?
(6) Azhar was getting late f or school, and had not yet had his breakf ast. While running out the
door, he picks up a f ruit f rom the f ruit basket which had the f ollowing f ruits. What is the
probability that Azhar picked a pear?
(7) Sehr was getting late f or his match. While running out the door, she picks up a f ruit f rom the
f ruit basket which had the f ollowing f ruits. What is the probability that Sehr did not pick a
mango?
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ID : pk-7-Data-Handling-Probability-Statistics [2]
Choose correct answer(s) f rom given choice
(8)
From a deck of cards, Asad withdraws a card at random. What is the probability that the card is
a Jack?
a.
1
11
c.
1
b.
52
1
d.
13
1
2
(9) T he average of any f our consecutive odd integers is always
a. even
b. a proper f raction
c. a prime number
d. a decimal number
(10) Ali got an average score of 80 in 4 tests. He got 81 as the average of the highest 3 scores, and
his lowest two scores are the same numbers. What is the average of his highest two scores?
a. 84
b. 82.5
c. 82
d. 83
(11) Mohammad measures the heights of 8 of his f riends as f ollows (in centimetres)
127 , 146 , 155 , 105.6 , 148 , 148 , 152 , 144
What is the average of their heights?
a. 140.7
b. 136.7
c. 145.7
d. 141.7
(12) T he table below shows the number of books read by f ive children in one month.
Name
Number of books
Shahzad
16
Amin
14
Adil
22
Inzamam
11
Rahman
22
What is the average number of books a child reads in a month?
a. 17
b. 23
c. 20
d. 16
(13) If a die is thrown, what is the probability of getting an odd number?
a.
4
b. 0
6
c.
1
6
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d.
1
2
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ID : pk-7-Data-Handling-Probability-Statistics [3]
(14) Find the mean of all the internal angles in a hexagon.
a. 110°
b. 100°
c. 140°
d. 120°
Fill in t he blanks
(15) T he average of Fahad's marks in 4 subjects is 81. He got 106 marks in 5th subject. His average
in all 5 subjects is
.
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ID : pk-7-Data-Handling-Probability-Statistics [4]
Answers
(1)
Pear
Step 1
Probability =
Number of f avorable outcomes
Number of possible equally-likely outcomes
Step 2
Since there are 12 f ruits in the basket, in which 5 are pears.
T heref ore the probability of a pear picked f rom the basket =
Number of pear in the basket
T otal number of f ruits in the basket
=
5
12
Step 3
If you look at the basket caref ully, you will notice that the number of pears in a basket are
more as compare to other f ruits. T heref ore the probability of picked a pear f rom the basket
is more as compare to other f ruits and hence we can say that the most likely f ruit to be
picked f rom the basket is pear.
(2)
16
52
Step 1
If you look at the question caref ully, you will notice that Sadia withdraw a card at random
f rom a deck of cards.
T otal number of cards in a deck = 52 cards
Prime number cards in a deck of card = {(2, 3, 5, 7) = Clubs,
(2, 3, 5, 7) = Diamonds,
(2, 3, 5, 7) = Hearts,
(2, 3, 5, 7) = Spades }
T otal prime number cards = 16
Step 2
Probability that the number on the card is a prime number =
T otal prime cards in a deck
T otal number of cards in a deck
=
16
52
Step 3
Now the probability that the number on the card is a prime number =
16
.
52
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ID : pk-7-Data-Handling-Probability-Statistics [5]
(3)
1
n3
Step 1
It is given that there are n numbers of which one is
T heref ore, the numbers are
and all the others are 1's.
, 1, 1, 1 …… (where n is the total number of
numbers in the series)
Step 2
Out of n numbers one is
and remaining n-1 numbers are 1.
T heref ore, the sum of n-1 numbers is n-1.
Now, the sum of all numbers in the series = n-1 + (1 -
1
n2
)=n-
1
n2
Step 3
Now, the arithmetic mean of the numbers =
Sum of the all numbers
n
=
1
n-
n2
n
=
n
-
n
=1-
1
n3
1
n3
Step 4
T hus, we can say that the arithmetic mean of these numbers is
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1
n3
less than 1.
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ID : pk-7-Data-Handling-Probability-Statistics [6]
(4) 11
Step 1
According to the question,
80C - 240A = 950 -----(1),
320A + 80B = 1690 -----(2)
Step 2
By adding equation (1) and (2), we get:
80C + 320A -240A + 80B = 2640
⇒ 80C + 80A + 80B = 2640
⇒ 80(C + A + B) = 2640
⇒ C + A + B = 2640/80
⇒ A + B + C = 33
Step 3
Now, the average of A, B, C =
A+ B+ C
3
=
33
3
= 11
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(5)
1
12
Step 1
Let's assume S, E and P are the Sample Space, Event and Probability that Ayaaz gets a
4 on the die and a T ail f rom the coin toss respectively.
S = { (1,T ), (2,T ), (3,T ), (4,T ), (5,T ), (6,T )
(1,H), (2,H), (3,H), (4,H), (5,H), (6,H) }
E = { (4 , T ) }
Step 2
P(E) =
n(E)
n(S)
(Since P(E) is the probability of getting event E, n(E) is the number of elements in the event
E and n(S) is the number of elements in the sample space S )
P(E) =
1
12
Step 3
Now, the probability that he gets a 4 on the die and a T ail f rom the coin toss is
1
.
12
(6)
7
22
Step 1
If you look at the basket caref ully, you will notice that there are 22 f ruits in the basket, in
which 7 are pear.
Step 2
Now the probability that Azhar picked a pear =
Number of pear in the basket
T otal number of f ruits in the basket
=
7
22
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(7)
15
21
Step 1
If you look at the basket caref ully, you will notice that there are 21 f ruits in the basket, in
which 6 are mango.
Step 2
T he probability that Sehr picked a mango =
Number of mango in the basket
T otal number of f ruits in the basket
=
6
21
Step 3
Now the probability that Sehr did not pick a mango = 1 - T he probability that Sehr picked a
mango
6
=1-
21
=
21 - 6
21
=
15
21
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(8)
1
c.
13
Step 1
If you look at the question caref ully, you will notice that Asad withdraws a card at random
f rom a deck of cards.
T otal number of cards in a deck = 52 cards
Number of Jack cards in a deck = 4 cards
Step 2
Number of Jack cards in a deck
T he probability that the card is a Jack =
T otal number of cards in a deck
4
=
52
1
=
13
Step 3
T heref ore, the probability of a Jack is
1
.
13
(9) a. even
Step 1
T he average of any f our consecutive odd integers is always even.
For example 1, 3, 5 and 7 are the f our consecutive odd integers,
average of 1, 3, 5 and 7 =
1+3+5+7
4
=
16
4
=4
Which is an even number.
Step 2
T heref ore we can say that the average of any f our consecutive odd integers is always an
even.
(10) d. 83
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(11) a. 140.7
Step 1
If you look at the question caref ully, you will notice that Mohammad measures the heights
of 8 of his f riends as f ollows (in centimetres)
127 , 146 , 155 , 105.6 , 148 , 148 , 152 , 144
Sum of heights of his f riends = 127 + 146 + 155 + 105.6 + 148 + 148 + 152 + 144 = 1125.6
T otal number of his f riends = 8
Step 2
Average height of Mohammad f riends =
Sum of heights of his f riends
T otal number of his f riends
=
1125.6
8
= 140.7
Step 3
Now the average height of Mohammad f riends is 140.7 .
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(12) a. 17
Step 1
If you look at the question caref ully, you will notice that the table below shows the number
of books read by f ive children in one month.
Name
Number of books
Shahzad
16
Amin
14
Adil
22
Inzamam
11
Rahman
22
T otal number of books read by children = 16 + 14 + 22 + 11 + 22 = 85
Step 2
Average number of books read by the children =
T otal number of books read by the children
number of children
=
16 + 14 + 22 + 11 + 22
5
=
85
5
= 17
Step 3
Now the average number of books read by the children is 17 .
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(13)
d.
1
2
Step 1
Let's assume S, E and P are the Sample Space, Event and Probability of getting an odd
number respectively.
S = {1, 2, 3, 4, 5, 6}
E = {1, 3, 5}
Step 2
P(E) =
n(E)
n(S)
(Since P(E) is the probability of getting event E, n(E) is the number of elements in the event
E and n(S) is the number of elements in the sample space S )
P(E) =
1
2
Step 3
Now, the probability of getting an odd number is
1
.
2
(14) d. 120°
Step 1
T he sum of all six internal angles in a hexagon must be 720°.
Now the mean of all the internal angles in a hexagon =
720
= 120
6
Step 2
T heref ore the mean of all the internal angles in a hexagon is 120°
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(15)
86
Step 1
We know that the average of marks in all subjects =
Sum of marks in all subjects
.
T otal number of subjects
T he average of Fahad's marks in 4 subjects =
Sum of marks in 4 subjects
= 81
4
⇒ Sum of marks in 4 subjects = 4 × 81 = 324.
Step 2
Since he got 106 marks in the 5th subject, the average of Fahad's marks in 5 subjects =
Sum of marks in 5 subjects
5
=
Sum of marks in 4 subjects + Marks in the 5th subject
5
=
324 + 106
5
=
430
5
= 86
Step 3
T heref ore, his average in all 5 subjects is 86.
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