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Published on: 13/05/2022
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Take MCQ Business Maths and Statistics Test

1.
The probability of the happening of an event X is 0.002 in an experiment. If an experiment is reported 1000 times, find the probability that the event X happens exactly twice? (e-2 = 0.1353)
2.
If you buy a lottery ticket in 50 lotteries, in each which your chance of winning a prize is \(\frac { 1 }{ 100 } \). What is the approximate probability that you will win a prize at least once (e-0.5 = 0.6066).
3.
The random variable X has the normal distribution f(x) = \(C{ e }^{ -\left( \frac { x-100 }{ 50 } \right) ^{ 2 } }\), then find the value of C.
4.
In a packet of 50 pens, 10 are defective, 10 pens are selected at random. What is the probability that atleast one is defective.
5.
Students of a class were given an aptitude test. Marks were found to be normally distributed with mean 60 and S.D. 5. Find the percentage of students who scored more than 60 marks.
1.
Let p be the probability of happening of an event.
Given p = 0.002 = \(\frac { 2 }{ 1000 } \)
Also n = 1000
∴ Mean = np = \(1000\times \frac { 2 }{ 1000 } \) = 2
Hence, X follows Poisson distribution with
P(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
∴ P( event happens exactly twice)
= P(X = 2)
= \(\frac { e^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { e^{ -2 }.{ (2 }^{ 2 }) }{ 2 } \)
= e-2(2) = 2(0.1353) = 0.2706
∴ P(X = 2) = 0.2706
2.
Let p be the probability of winning the prize
Given p = \(\\ \frac { 1 }{ 100 } \) and n = 50
∴ Mean = np = \(\frac { 1 }{ 100 } \times 50=\frac { 1 }{ 2 } \)
∴ λ = 0.5
Hence, X follows Poisson distribution with
P(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
∴ P(winning the prize at least once) = P(X ≥ 1)
= 1-P(X < 1)
= 1-P(X = 0)
= 1-\(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \) = 1-e-0.5
= 1-0.6066
∴ P(X≥1) = 0.3934
3.
The probability function for the normal distribution is
f(x) = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) ^{ 2 } }\), -∞
= \(C{ e }^{ -\frac { 1 }{ 2 } \left( \frac { x-100 }{ 25 } \right) ^{ 2 } }\)
= \(C{ e }^{ -\frac { 1 }{ 2 } \left( \frac { x-100 }{ 5 } \right) ^{ 2 } }\) ...(2)
Comparing (1) and (2), μ =100, σ = 5 and
C = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } =\frac { 1 }{ 5\sqrt { 2\pi } } \)
∴ C = \(\frac { 1 }{ 5\sqrt { 2\pi } } \).
4.
Given n = 10
Probability of selecting a defective pen = p
= \(\frac { 10 }{ 50 } =\frac { 1 }{ 5 } \)
q = 1-p = \(1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
∴ P(X = x) = nCx pxqn-x
P (at least one pen is defective)
= P(X≥1) = 1-P(X<1)
= 1-P(X = 0)
= 1-10C0 \(\left( \frac { 1 }{ 5 } \right) ^{ 0 }\left( \frac { 4 }{ 5 } \right) ^{ 10 }\)
= 1-\(\frac { { 4 }^{ 10 } }{ { 5 }^{ 10 } } \)
5.
Given mean μ = 60 and S.D. σ = 5
To find P(X > 60)
When X = 60, Z =\(\frac { X-\mu }{ \sigma } =\frac { 60-60 }{ 5 } \) = 0
∴ P(X > 60) = P(Z > 0) = P (0 < Z < ∞)
= 0.5
∴ 50% of students scored more than 60 marks
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