Probability and Statistics MCQ Quiz - Objective Question with Answer for Probability and Statistics - Download Free PDF

Explanation:

Given data a, a + d, a + 2d, ......... a + 2nd

Mean (\(\bar x\)) = sum of observations / number of observations

         = \( a+ a + d+a + 2d+ ...+ a + 2nd\over 2n+1\)

       = \( a(2n+1)+d(1+2+3+..+2n)\over 2n+1\)

      = \( a(2n+1)+d{2n(2n+1)\over 2}\over 2n+1\)

     = \( a(2n+1)+d{n(2n+1)}\over 2n+1\) = a + nd

Now, deviations from mean = x - \(\bar x\)

 = nd, (n - 1)d, (n - 2)d, ...., 0, d, 2d, ..., (n-1)d, nd

So, mean deviation from mean

= \(nd+ (n - 1)d+(n - 2)d+ ...+ 0+ d+ 2d+ ...+ (n-1)d+ nd\over 2n+1\)

= \(2d(1+2+...+n)\over 2n+1\)

= \(2d{n(n+1)\over 2}\over 2n+1\) = \(\rm \frac{n(n+1)}{(2n+1)}d\)

Option (2) is true.

Calculation:

Here,  5, 8, 9, 5, 6, 4, 9, 3, 9, 3, 6, 1, 9, 7, 1, 2, 9 and 5

Here,

9 comes 5 times

∴ Mode of given data is 9.

Formula

Standard deviation = σ = √[1/N(∑f_i(x_i – x̅)²]

x̅ = Mean

x̅ = ∑fx/N

N = Total number of frequency

Calculation

x̅ = ∑fx/N                                                                                                                        

⇒ (175/10)

∴ x̅ = 17.5

⇒ Standard deviation = √(1/10)[170]

∴ Standard deviation is √17

Formula

Mean = ∑x_if_i/n

f = frequency

x = observatons

n = number of observations

Calculation

Mean = 83/14

∴ Mean of this series is 83/14

The arithmetic mean is denotted by X̅ is given by

X̅ = (x1₁ + x₂ + ------ x_n)/n

X̅ = ∑x_i/n

Where as (x₁ + x₂ + ------ x_n) are observations

n = Number of observation

1 - For any distribution the sum of deviation from the arithmetic mean is always zero

2 - If each value of the variate x is increased or decreased by a constant value then the arithmetic mean of the variate so obtained is also increased or decreased by the same constant value

3 - The sum of the square of the deviation of a set of values is minimum when taken about mean.

4 - If the values of the variate are multiplied or divided by a constant value, the arithmetic mean so obtained is same as the initial arithmetic mean multiplied ro divided by the constant value

Given

Mean of certain number = 20

Mean of square of same observation = 500

Formula used

σ = √[(∑x²/n –(∑x/n)²]

(∑x/n) = mean

∑x²/n = mean of square

Calculation

Standard deviation is given by

⇒ √[(∑(x²/n –(∑x/n)²] = √[500 – (20)²]

⇒ √(500 – 400)

∴ Standard deviation is 10

Given:

The data = 30, 34, 35, 36, 37, 38, 39, 40 

Formula used:

The median for odd number = \(({n\ +\ 1\over 2})^{th}\)

The median for even number = \({({n\over 2})^{th}\ +\ ({{n\over 2}\ +\ 1})^{th}}\over 2\)     Where, n = The total number of terms (odd or even)

Calculation:

Let us assume the median of the even number be X

⇒ The total even number = 8

⇒ Then the value of n = 8

⇒ The median of the even number = \({({8\over 2})^{th}\ +\ ({{8\over 2}\ +\ 1})^{th}}\over 2\) = (4 + 5)/2 = (36 + 37)/2 = 36.5

⇒ When 35 removed then total number = n = 7

⇒ The median of the data when 35 is removed = \(({7\ +\ 1\over 2})^{th}\) = 8/2 = 4th term = 37

⇒ The difference of the median = 37 - 36.5 = 0.5

∴ The required result will be 0.5.

⇒ Number of coins having tail on both sides = n

⇒ Number of fair coins = n + 1

ATQ,

⇒ Probability of getting a tail = 31/42

⇒ P(tail) = \(\frac{{{}_{}^n{C_1}}}{{{}_{}^{2n\; + \;1}{C_1}}} \times 1 + \frac{{{}_{}^{n\; + \;1}{C_1}}}{{{}_{}^{2n\; + \;1}{C_1}}} \times \frac{1}{2} = 31/42\) 

\( \Rightarrow \frac{n}{{2n\; + \;1}} + \frac{{n\; + \;1}}{{2\left( {2n\; + \;1} \right)\;}} = 31/42\)

⇒ (3n + 1) x 21 = 31(2n + 1)

⇒ 63n + 21 = 62n + 31

⇒ n = 10

∴ Total coins in bag = 2n + 1 = 21

Concept:

As per the given data,

9, 5, 8, 9, 9, 7, 8, 9, 8

Arranging the numbers in numerical order

5, 7, 8, 8, 8, 9, 9, 9, 9

Since there is an odd number of numbers, the middle number is the median of the given data

⇒ Median = 8

The value which appears mostly is considered as mode, as 9 is repeated 4 times.

⇒ Mode = 9

Mean = (9 + 5 + 8 + 9 + 9 + 7 + 8 + 9 + 8)/9 = 8

∴ Median, mode and mean = (8, 9, 8)

Given

P(A) = 2P(B) = 6P(C)

Concept

When event are mutually exclusive and exhaustive

P(A) + P(B) + P(C) = 1

Calculation

Let P(A) be k

⇒ P(B) = k/2

⇒ P(C) = k/6

So according to the concept

⇒ k + k/2 + k/6 = 1

⇒ 10k/6 = 1

⇒ k = 3/5

Concept:

For a continuous random variable, f(x) is called probability density function if it satisfies-

​\(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right)dx = 1\)

Relation between Probability density function f(x) and Probability distribution function is:

\(f\left( x \right) = \frac{d}{{dx}}\left\{ {F\left( x \right)} \right\}\)

Calculation:

Here Probability distribution function F(x) is given 

\(\because \mathop \smallint \limits_{ - \infty }^\infty f\left( x \right)dx = 1\)

\( \Rightarrow \mathop \smallint \limits_{ - \infty }^1 f\left( x \right)dx + \mathop \smallint \limits_1^3 f\left( x \right)dx + \mathop \smallint \limits_3^\infty f\left( x \right)dx = 1\)

\( \Rightarrow \mathop \smallint \limits_{ - \infty }^1 \frac{d}{{dx}}\left\{ {F\left( x \right)} \right\}dx + \mathop \smallint \limits_1^3 \frac{d}{{dx}}\left\{ {F\left( x \right)} \right\}dx + \mathop \smallint \limits_3^\infty \frac{d}{{dx}}\left\{ {F\left( x \right)} \right\}dx = 1\)

\( \Rightarrow \mathop \smallint \limits_{ - \infty }^1 \frac{d}{{dx}}\left\{ 0 \right\}dx + \mathop \smallint \limits_1^3 \frac{d}{{dx}}\left\{ {k{{\left( {x - 1} \right)}^4}} \right\}dx + \mathop \smallint \limits_3^\infty \frac{d}{{dx}}\left\{ 1 \right\}dx = 1\)

\( \Rightarrow 0 + \mathop \smallint \limits_1^3 4k{\left( {x - 1} \right)^3}dx +0= 1\)

\( \Rightarrow 4\frac{k}{4}\left[ {{{\left( {x - 1} \right)}^4}} \right]_1^3 = 1\)

\( \Rightarrow k\left[ {{{\left( {3 - 1} \right)}^4}} \right] = 1\)

\( \Rightarrow k = \frac{1}{{16}}\)

Explanation:

A standard normal distribution having zero mean and unit variance. A standard normal curve has 99.7% area in limits -3 to +3

Note: A standard normal curve has 68% area in limits -1 to +1 and 95% area in limits -2 to +2.

These are the standard results.

Explanation:

Number of ways in which a king can be drawn from the pack of 52 cards is \({}_{}^4{C_1}\) since there are 4 kings in a deck of cards

Similarly, a queen and a jack can be drawn in \({}_{}^4{C_1}\) ways

∴ Number of ways in which a king, a queen and a jack can be drawn is \(= {}_{}^4{C_1} \times {}_{}^4{C_1} \times {}_{}^4{C_1}\)

Number of ways in which three cards can be drawn from the pack of 52 cards is

\(= {}_{}^{52}{C_3}\)

\(\therefore The\;required\;probability = \frac{{sample\;space\;}}{{Total\;possible\;ways}}\)

\(\therefore The\;required\;probability = \frac{{{}_{}^4{C_1} \times {}_{}^4{C_1} \times {}_{}^4{C_1}}}{{{}_{}^{52}{C_3}}}\)

\(= \frac{{4 \times 4 \times 4}}{{\frac{{52 \times 51 \times 50}}{6}}}\)

\(= \frac{{16}}{{5525}}\)

\(P\left( {X \le \frac{2}{3}{\rm{|}}X > \frac{1}{3}} \right) = \frac{{P\left( {X \le \frac{2}{3} \cap X > \frac{1}{3}} \right)}}{{P\left( {X > \frac{1}{3}} \right)}}\)

\( = \frac{{P\left( {\frac{1}{3} < X \le \frac{2}{3}} \right)}}{{P\left( {X > \frac{1}{3}} \right)}}\)

\(P\left( {\frac{1}{3} < X \le \frac{2}{3}} \right) = \mathop \smallint \limits_{1/3}^{2/3} 4{x^3}dx = \left[ {{x^4}} \right]_{\frac{1}{3}}^{\frac{2}{3}} = \frac{{15}}{{81}} = \frac{5}{{27}}\)

\(P\left( {X > \frac{1}{3}} \right) = \mathop \smallint \limits_{1/3}^1 4{x^3}dx\;\)

\( = \left[ {{x^4}} \right]_{\frac{1}{3}}^1\)

\( = 1 - \frac{1}{{81}} = \frac{{80}}{{81}}\)

\(P\left( {X \le \frac{2}{3}{\rm{|}}X > \frac{1}{3}} \right) = \frac{{\frac{5}{{27}}}}{{\frac{{80}}{{81}}}} = \frac{5}{{27}} \times \frac{{81}}{{80}} = \frac{3}{{16}}\)

Given

In Poisson distribution

P(X = 0) = 0.6

Formula

Poisson distribution is given by

f(x) = e^-λλ^x/x!

Calculation

P(X = 0) = e^-λλ⁰/0!

⇒ 0.6 = e^-λ

⇒ 1/e^λ = 6/10 = 3/5

⇒ e^λ = 5/3

Taking log on both side

⇒ log_ee^λ = log_e(5/3)

∴ λ = Log_e(5/3)

Let λ be the parameter of exponential distribution

So, for exponentially distributed random variables X & Y

Expected Value or Mean \(E\left( X \right) = \frac{1}{\lambda }\)

Variance: \(Var\;\left( X \right) = \frac{1}{{{\lambda ^2}}}\)

Var (X + Y) = Var X + Var Y + 2 Cov (X, Y)

Where Cov (X, Y) = Covariance of X and Y

Correction coefficient is given by:

\(r = \frac{{Cov\;\left( {X,\;Y} \right)}}{{\sqrt {Var\left( X \right)Var\left( Y \right)} }}\)

Calculation:

Given: E(X) = E(Y) = 0.5

\(\frac{1}{{{\lambda _1}}} = \frac{1}{{{\lambda _2}}} = 0.5\)

λ₁ = λ₂ = 2

\(Var\;\left( X \right) = \frac{1}{{\lambda _1^2}} = \frac{1}{{{2^2}}} = 0.25\)

\(Var\;\left( Y \right) = \frac{1}{{\lambda _2^2}} = \frac{1}{{{2^2}}} = 0.25\)

Also it is given, Z = X + Y

And Var (Z) = 0

∴ Var (X + Y) = 0

Var X + Var Y + 2 Cov (X, Y) = 0

On putting values:

0.25 + 0.25 + 2 Cov (X, Y) = 0

\(Cov\;\left( {X,\;Y} \right) = - \frac{{0.5}}{2} = - 0.25\)

Correction coefficient is given by: