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Probability Distribution

Long Questions (Question 10)

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Question 10:
(a) It is found that 60% of the students from a certain class obtained grade A in English in O level trial examination.
If 10 students from the class are selected at random, find the probability that
(i) exactly 7 students obtained grade A.
(ii) not more than 7 students obtained grade A.

(b) Diagram below shows a standard normal distribution graph representing the volume of soy sauce in bottles produced by a factory.

It is given the mean is 950 cm3 and the variance is 256 cm6. If the percentage of the volume more than V is 30.5%, find
(i) the value of V,
(ii) the probability that the volume between 930 cm3 and 960 cm3.

Solution:
(a)(i) P(X=r)= c n r . p r . q n−r P(X=7)= C 10 7 ( 0.6 ) 7 ( 0.4 ) 3    =0.0860 ( ii ) P(X≤7) =1−P(X>7) =1−P( X=8 )−P( X=9 )−P( X=10 ) =1− C 10 8 ( 0.6 ) 8 ( 0.4 ) 2 − C 10 9 ( 0.6 ) 9 ( 0.4 ) 1 − C 10 10 ( 0.6 ) 10 ( 0.4 ) 0 =1−0.1209−0.0403−0.0060 =0.8328

(b)( i ) P( X>V )=30.5% P( Z> V−950 16 )=0.305 P( Z>0.51 )=0.305    V−950 16 =0.51 V=0.51( 16 )+950    =958.16  cm 3

( ii ) Probability =P( 930<X<960 ) =P( 930−950 16 <Z< 960−950 16 ) =P( −1.25<Z<0.625 ) =1−P( Z>1.25 )−P( Z>0.625 ) =1−0.1056−0.2660 =0.6284

Long Questions (Question 9)

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Question 9:
(a) 30% of the pens in a box are blue. Charlie picks 4 pens at random. Find the probability that at least one pen picked is not blue.

(b) The mass of papayas harvested from an orchard farm follows a normal distribution with a mean of 2 kg and a standard deviation of h kg. It is given that 15.87% of the papayas have a mass more than 2.5 kg.
(i) Calculate the value of h.
(ii) Given the number of papayas harvested from the orchard farm is 1320, find the number of papayas that have the mass between 1.0 kg and 2.5 kg.

Solution:
 (a) P( X≥1 )=1−P( X=0 )               =1− C 4 4 ( 0.3 ) 4 ( 0.7 ) 0               =0.9919

(b) μ=2, σ=h ( i ) P( X>2.5 )=15.87% P( Z> 2.5−2 h )=0.1587 P( Z>1.0 )=0.1587        2.5−2 h =1.0                 h=0.5


( ii ) p=P( 1.0<x<2.5 )   =P( 1.0−2 0.5 <Z< 2.5−2 0.5 )   =P( −2<Z<1 )   =1−P( Z<−2 )−P( Z>1 )   =1−P( Z>2 )−P( Z>1 )   =1−0.0228−0.1587   =0.8185 Number of papayas=0.8185×1320   =1080

Long Questions (Question 7)

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Question 7:
In a boarding school entry exam, 300 students sat for a mathematics test. The marks obtained follow a normal distribution with a mean of 56 and a standard deviation of 8.

(a) Find the number of students who pass the test if the passing mark is 40.

(b) If 12% of the students pass the test with grade A, find the minimum mark to obtain grade A.

Solution:
Let X=marks obtained by students X~N( 56, 8 2 ) ( a ) P( X≥40 )=P( Z≥ 40−56 8 )    =P( Z≥−2 )    =1−P( Z≥2 )    =1−0.02275    =0.9773 Number of students who pass the test =0.9773×300 =293 ( b ) Let the minimum mark to obtain grade A be k P( X≥k )=0.12 P( Z≥ k−56 8 )=0.12   k−56 8 =1.17   k=( 1.17 )( 8 )+56 =65.36

Thus, the minimum mark to obtain grade A is 66.

Long Questions (Question 6)

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Question 6:
The masses of tomatoes in a farm are normally distributed with a mean of 130 g and standard deviation of 16 g. Tomato with weight more than 150 g is classified as grade ‘A’.

(a) A tomato is chosen at random from the farm. Find the probability that the tomato has a weight between 114 g and 150 g. 

(b) It is found that 132 tomatoes in this farm are grade ‘A’. Find the total number of tomatoes in the farm.

Solution:
µ = 130
σ = 16

(a)
P ( 114 < X < 150 ) = P ( 114 − 130 16 < Z < 150 − 130 16 ) = P ( − 1 < Z < 1.25 ) = 1 − P ( Z > 1 ) − P ( Z > 1.25 ) = 1 − 0.1587 − 0.1056 = 0.7357


(b)
Probability of getting grade ‘A’ tomatoes,
P (X > 150) = P (Z > 1.25)
= 0.1056
Lets total number of tomatoes = N 0.1056 × N = 132 N = 132 0.1056 N = 1250

Long Questions (Question 3 & 4)

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Question 3:
The result of a study shows that 20% of students failed the Form 5 examination in a school. If 8 students from the school are chosen at random, calculate the probability that
(a) exactly 2 of them who failed,
(b) less than 3 of them who failed.

Solution:
(a)
p = 20% = 0.2,
q = 1 – 0.2 = 0.8
X ~ B (8, 0.2)
P (X = 2)
=  C 8 2  (0.2)2 (0.8)6
= 0.2936

(b)
P (X < 3)
= P (X = 0) + P (X = 1) + P (X = 2)
= C 8 0  (0.2)0(0.8)8 +   C 8 1  (0.2)1(0.8)7 +   C 8 2  (0.2)2(0.8)6
= 0.16777 + 0.33554 + 0.29360
= 0.79691



Question 4:
In a survey carried out in a particular district, it is found that three out of five families own a LCD television.
If 10 families are chosen at random from the district, calculate the probability that at least 8 families own a LCD television.

Solution:
Let X be the random variable representing the number of families who own a LCD television. X~B( n,p ) X~B( 10,  3 5 ) p= 3 5 =0.6 q=1−0.6=0.4 P(X=r)= c n r . p r . q n−r P( at least 8 families own a LCD television ) P(X≥8) =P( X=8 )+P( X=9 )+P( X=10 ) = C 10 8 ( 0.6 ) 8 ( 0.4 ) 2 + C 10 9 ( 0.6 ) 9 ( 0.4 ) 1 + C 10 10 ( 0.6 ) 10 ( 0.4 ) 0 =0.1209+0.0403+0.0060 =0.1672

8.2c Probability Of An Event

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Example:

The masses of pears in a fruit stall are normally distributed with a mean of 220g and a variance of 100g. Find the probability that a pear that is picked at random has a mass

(a) of more than 230g.
(b) between 210g and 225g.

Hence, find the value of h such that 90% of the pears weigh more than h g.

Solution:

m= 220g
σ = √100 = 10g
Let X be the mass of a pear.

(a)

P( X>230 ) =P( Z> 230−220 10 )← Convert to standard normal distribution using Z= X−μ σ =P( Z>1 ) =0.1587

(b)

P( 210<X<225 ) =P( 210−220 10 <Z< 225−220 10 )← Convert to standard normal distribution using Z= X−μ σ =P( −1<Z<0.5 ) =1−P( Z>1 )−P( Z>0.5 ) =1−0.1587−0.3085 =0.5328

For 90% (probability = 0.9) of the pears weigh more than h g, 
P(X > h) = 0.9
P(X < h) = 1 – 0.9
                = 0.1
From the standard normal distribution table,
P(Z > 0.4602) = 0.1
P(Z < –0.4602) = 0.1

h−220 10 =−0.4602 h−220=−4.602 h=215.4

 

 

 

8.1a Probability Of An Event That Follows A Binomial Distribution

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8.1a Probability of an Event that follows a Binomial Distribution
 
1. For a Binomial Distribution, the probability of obtaining r numbers of successes out of n experiments is given by

P ( X = r ) = c n r . p r . q n − r

where
P = probability
X = discrete random variable
r =  number of success (0, 1, 2, 3, …, n)
n = number of trials
p = probability of success in an experiment (0 < p <1)
q = probability of failure in an experiment (q = 1 – p)



Example 1
Kelvin has taken 3 shots in a shooting practice. The probability that Kelvin strikes the target is 0.6. X represents the number of times kelvin strikes the target.

(a)   List the elements of the binomial discrete random variable X.

(b) Calculate the probability for the occurrence of each of the elements of X.

(c)  Hence, plot a graph to represent the binomial probability distribution of X.
 
Solution:
(a) X = Number of times Kelvin strikes the target
 X = {0, 1, 2, 3}
 
(b) X ~ B (n, p)
  X ~ B (3, 0.6)

P ( X = r ) = c n r . p r . q n − r ( i ) P ( X = 0 ) = C 3 0 ( 0.6 ) 0 ( 0.4 ) 3 ← Probability of failure = 1 − 0.6 = 0.4 = 0.064 ( i i ) P ( X = 1 ) = C 3 1 ( 0.6 ) 1 ( 0.4 ) 2 = 0.288 ( i i i ) P ( X = 2 ) = C 3 2 ( 0.6 ) 2 ( 0.4 ) 1 = 0.432 ( i v ) P ( X = 3 ) = C 3 3 ( 0.6 ) 3 ( 0.4 ) 0 = 0.216


(c)


Long Questions (Question 5)

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Question 5:
The diameter of oranges harvested from a fruit orchard has a normal distribution with a mean of 3.2 cm and a variance of 2.25 cm.
Calculate
(a) the probability that an orange chosen at random from this fruit orchard has a diameter of more than 3.8 cm.
(b) the value of k if 30.5 % of the oranges have diameter less than k cm.

Solution:
µ = 3.2 cm
σ2 = 2.25cm
σ = √2.25 = 1.5 cm
Let X represents the diameter of an orange.
X ~ N (3.2, 1.52)

(a)
P ( X > 3.8 ) = P ( Z > 3.8 − 3.2 1.5 ) = P ( Z > 0.4 ) = 0.3446

(b)
P ( X < k ) = 0.305 P ( Z < k − 3.2 1.5 ) = 0.305 From the standard normal distribution table, P ( Z > 0.51 ) = 0.305 P ( Z < − 0.51 ) = 0.305 k − 3.2 1.5 = − 0.51 k − 3.2 = − 0.765 k = 2.435


Long Questions (Question 8)

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Question 8:
(a) A survey is carried out about red crescents in a school.
It is found that the mean of the number of red crescents is 315, the variance is 126 and the probability that a student participate in red crescents is p.
(i) Find the value of p.
(ii) If 8 students from the school are chosen at random, find the probability that more than 5 students participate in red crescents.

(b) The mass of fertiliser used in an orchard farm is normally distributed with mean 5 kg and variance 0.8 kg. Find the probability that on a particular day, more than 6 kg of fertiliser is used.

Solution:
(a)(i) np=315 np( 1−p )=126 315( 1−p )=126  1−p= 126 315  1−p=0.4    p=0.6 ( ii ) n=8,p=0.6 P(X=r)= c n r . p r . q n−r P(X=r)= C 8 r ( 0.6 ) r ( 0.4 ) 8 P(X>5) =P( X=6 )+P( X=7 )+P( X=8 ) = C 8 6 ( 0.6 ) 6 ( 0.4 ) 2 + C 8 7 ( 0.6 ) 7 ( 0.4 ) 1 + C 8 8 ( 0.6 ) 8 ( 0.4 ) 0 =0.20902+0.08958+0.01680 =0.3154

(b) P( X>6 )=P( Z> 6−5 0.8 )    =P( Z>1.12 )    =0.1314

Short Questions (Question 3 & 4)

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Question 3:
The masses of mangoes in a stall have a normal distribution with a mean of 200 g and a standard deviation of 30 g.
(a) Find the mass, in g, of a mango whose z-score is 0.5.
(b) If a mango is chosen at random, find the probability that the mango has a mass of at least 194 g.

Solution:
µ = 200 g
σ = 30 g
Let X be the mass of a mango.

(a)
X − 200 30 = 0.5 X = 0.5 ( 30 ) + 200 X = 215 g

(b)
P ( X ≥ 194 ) = P ( Z ≥ 194 − 200 30 ) = P ( Z ≥ − 0.2 ) = 1 − P ( Z > 0.2 ) = 1 − 0.4207 = 0.5793



Question 4:
Diagram below shows a standard normal distribution graph.


The probability represented by the area of the shaded region is 0.3238.
(a) Find the value of k.
(b) X is a continuous random variable which is normally distributed with a mean of 80 and variance of 9.
Find the value of X when the z-score is k.

Solution:
(a)
P(Z > k) = 0.5 – 0.3238 
= 0.1762
k = 0.93

(b)
µ = 80,
σ2 = 9, σ = 3
X − 80 3 = 0.93 X = 3 ( 0.93 ) + 80 X = 82.79