Test: Statistics Part:- 2 - Mathematics MCQ

# Test: Statistics Part:- 2 - Mathematics MCQ

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## 10 Questions MCQ Test Additional Topics for IIT JAM Mathematics - Test: Statistics Part:- 2

Test: Statistics Part:- 2 for Mathematics 2024 is part of Additional Topics for IIT JAM Mathematics preparation. The Test: Statistics Part:- 2 questions and answers have been prepared according to the Mathematics exam syllabus.The Test: Statistics Part:- 2 MCQs are made for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Statistics Part:- 2 below.
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Test: Statistics Part:- 2 - Question 1

### Suppose we uniformly and randomly select a permutation from the 20! permutations of 1, 2, 3 ….., 20. What is the probability that 2 appears at an earlier position that any other even number in the selected permutation?

Detailed Solution for Test: Statistics Part:- 2 - Question 1

Number of permutations with ‘2’ in the first position = 19!
Number of permutations with ‘2’ in the second position = 10 × 18!
(fill the first space with any of the 10 odd numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways)
Number of permutations with ‘2’ in 3rd position = 10 × 9 × 17!
(fill the first 2 places with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers)
and so on until ‘2’ is in 11th place. After that it is not possible to satisfy the given condition, since there are only 10 odd numbers available to fill before the ‘2’. So the desired number of permutations which satisfies the given condition is
19! + 10 x 18! + 10 x 9 x 17! + 10 x 9 x 8 x 16!+ .... + 10! x 9!
Now the probability of this happening is given by

Which is clearly not choices (a),(b) or (c)
Thus, Answer is (d) none of these.

Test: Statistics Part:- 2 - Question 2

### Aishwarya studies either computer science or mathematics everyday. if the studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?

Detailed Solution for Test: Statistics Part:- 2 - Question 2

Let C denote computes science study and  M denotes maths study.
P(C on monday and C on wednesday) + p(C on monday, c on tuesday and C on wednesday)
= 1 x 0.6 x 0.4 + 1 x 0.4 x 0.4
= 0.24 + 0.16
= 0.40

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Test: Statistics Part:- 2 - Question 3

### Let X be a randon variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean –1 and variance unknown. If P(X ≤ –1) = P(Y ≥ 2) the standard deviation of Y is

Detailed Solution for Test: Statistics Part:- 2 - Question 3

given
Convert into standard and normal variates,

...(i)
Now since us know that in standard normal distribution,

Test: Statistics Part:- 2 - Question 4

An unbalanced ice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same.
If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closed to the probability that the face value exceeds 3?

Detailed Solution for Test: Statistics Part:- 2 - Question 4

It is given that
P(odd) = 0.9 p(even)
Now since ∑p(x) = 1
∴ p(odd) + p(even) = 1
⇒ 0.9p(even) + p(even) = 1
⇒ p(even) = 1/1.9 = 0.5263
Now, it is given that p (any even face) is same
i.e p(2) = p(4) = p(6)
Now since,

∴
= 1/3 (0.5263)
= 0.1754
It is given that

Test: Statistics Part:- 2 - Question 5

Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company, therefore, subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q.
What is the probability of a computer being declared faulty?

Detailed Solution for Test: Statistics Part:- 2 - Question 5

From the diagram,
Pdeclared faulty = pq + (1 − p) (1 − q )

Test: Statistics Part:- 2 - Question 6

What is the probability that a divisor of 1099 is a multiple of 1096?

Detailed Solution for Test: Statistics Part:- 2 - Question 6

p(multiple of 10% |divisor of 1099)

Since 10 = 2.5
1099 = 299 . 599
Any divisor of 1099 is of the form 2a . 5b where 0 ≤ a ≤ 99 and 0 ≤ b ≤ 99.
The number of such possibilities is combination of 100 values of a and 100 values of b = 100 × 100 each of which is a divisor of 1099
So, no. of divisors of 1099 = 100 × 100.
Any number which is a multiple of 1096 as well as divisor of 1099 is of the form 2a . 5b where 96 ≤  a ≤  99 and 96 ≤ b ≤  99. The number of such combinations of 4 values of a and 4 values of b is 4 × 4 combinations, each of which will be a multiple of 1096 as well as a divisor of 1099
∴ p(multiple of 1096|divisor of 1099

Test: Statistics Part:- 2 - Question 7

Let P(E) denote the probability of the even E. Given P(A) = 1, P(B) = 1/2, the values of P(A/B) and p(B/A) respectively are

Detailed Solution for Test: Statistics Part:- 2 - Question 7

Here, P(A) = 1, P(B) = 1/2
Since A, B are independent events,
∴ P(AB) = P(A)P(B)

Test: Statistics Part:- 2 - Question 8

A program consists of two modules executed sequentially. Let f1 (t) and f2 (t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by

Detailed Solution for Test: Statistics Part:- 2 - Question 8

Let the time taken for first and second modules be represented by x and y and
total time = t.
and y and total time = t.
∴ t = x + y is a random variable
Now the joint density function

which is also called as convolution of f1 and f2, abbreviated as f1* f2.

Test: Statistics Part:- 2 - Question 9

If a fair coin is tossed four times. What is the probability that two heads and two tails will result?

Detailed Solution for Test: Statistics Part:- 2 - Question 9

Here P(H) = P(T) = 1/2
It's a Bernoulli's trials.
∴ Required probability

Test: Statistics Part:- 2 - Question 10

An examination paper has 150 multiple-choice questions of one mark each, with each question having four choices. Each incorrect answer fetches – 0.25 mark. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained all these students is

Detailed Solution for Test: Statistics Part:- 2 - Question 10

Let the marks obtained per question be a random variable X. It’s probability  distribution table is given below:

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