Probability and Statistics Questions and Answers – Random Variables

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This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Random Variables”.

1. Consider a dice with the property that that probability of a face with n dots showing up is proportional to n. The probability of face showing 4 dots is?
a) \(\frac{1}{7} \)
b) \(\frac{5}{42} \)
c) \(\frac{1}{21} \)
d) \(\frac{4}{21} \)
View Answer

Answer: d
Explanation: P (n) is proportional to n where n=
1,2,3,…6 is random variable.
P(n) = kn
P(1)+P(2)….P (6) = 1
K(1+2+3+4+5+6) = 1
\(K=\frac{1}{21} \)
Hence P(4) = 4K = \(\frac{4}{21}. \)
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2. Let X be a random variable with probability distribution function f (x)=0.2 for |x|<1
= 0.1 for 1 < |x| < 4
= 0 otherwise
The probability P (0.5 < x < 5) is _____
a) 0.3
b) 0.5
c) 0.4
d) 0.8
View Answer

Answer: c
Explanation: P (0.5 < x < 5) = Integrating f (x) from
0.5 to 5 by splitting in 3 parts that is from 0.5 to 1
and from 1 to 4 and 4 to 5 we get
P (0.5 < x < 5) = 0.1 + 0.3 + 0
P (0.5 < x < 5) = 0.4.

3. Runs scored by batsman in 5 one day matches are 50, 70, 82, 93, and 20. The standard deviation is ______
a) 25.79
b) 25.49
c) 25.29
d) 25.69
View Answer

Answer: a
Explanation: The mean of 5 innings is
(50+70+82+93+20)÷5 = 63
S.D = [1n (x(n)-mean)2]0.5
S.D = 25.79.

4. Find median and mode of the messages received on 9 consecutive days 15, 11, 9, 5, 18, 4, 15, 13, 17.
a) 13, 6
b) 13, 18
c) 18, 15
d) 15, 16
View Answer

Answer: b
Explanation: Arranging the terms in ascending order 4, 5, 9, 11, 13, 14, 15, 18, 18.
Median is \(\frac{(n+1)}{2} \) term as n = 9 (odd) = 13.
Mode = 18 which is repeated twice.

5. Mode is the value of x where f(x) is a maximum if X is continuous.
a) True
b) False
View Answer

Answer: a
Explanation: For a continuous variable mode is defined as the value where f(x) is a maximum or it is defined as the quantity repeated maximum number of times.
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6. E (XY)=E (X)E (Y) if x and y are independent.
a) True
b) False
View Answer

Answer: a
Explanation: By the property of Expectation
E (XY) = E (X) E (Y).
That is the Expectation of a composite function XY is the product of the individual expectations of X and Y.

7. A coin is tossed up 4 times. The probability that tails turn up in 3 cases is ______
a) \(\frac{1}{2} \)
b) \(\frac{1}{3} \)
c) \(\frac{1}{4} \)
d) \(\frac{1}{6} \)
View Answer

Answer: a
Explanation: p=0.5 (Probability of tail)
q=1-0.5=0.5
n=4 and x is binomial variate.
P (X=x) = nCx px qn-x.
P (X=3) = 4C3 (0.5)3 = 12.

8. If E denotes the expectation the variance of a random variable X is denoted as?
a) (E(X))2
b) E(X2)-(E(X))2
c) E(X2)
d) 2E(X)
View Answer

Answer: b
Explanation: By property of Expectation
V (X) = E (X2)-(E(X))2.

9. X is a variate between 0 and 3. The value of E(X2) is ______
a) 8
b) 7
c) 27
d) 9
View Answer

Answer: d
Explanation: Integrating f(x) = x2 from 0 to 3 we get E(X2) = 32 = 9.
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10. The random variables X and Y have variances 0.2 and 0.5 respectively. Let Z= 5X-2Y. The variance of Z is?
a) 3
b) 4
c) 5
d) 7
View Answer

Answer: d
Explanation: Var(X) = 0.2, Var(Y) = 0.5
Z = 5X – 2Y
Var(Z) = Var(5X-2Y)
= Var(5X) + Var(2Y)
= 25Var(X) + 4Var(Y)
Var(Z) = 7.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn