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Questions:

Generic 3

15 x 3 = 45

The average percentage of failure in a certain examination is 40. What is the probàbility that out of a group of 6 candidates atleast 4 passed in the examination?

In a binomial distribution consisting of 5 independent trails, probability of 1 and 2 Successes are 0.4096 and 0.2048 respectively. Find p.

Ten coins are thrown simultaneously. Find the probability of getting atleast 7 heads.

On an average if one vessel in every ten is Wrinkled, find the probability that out of 5 vessels expected to arrive, atleast 4 will arrive safely.

If a publisher of non-technical books takes a great pain to ensure that his books are fire of typological errors, so that the probability of any given page containing atleast one such error is 0.005 and errors are independent from page to page
(i) what is the probability that one of its 400 page novels will contain exactly one page with error
(ii) atmost 3 pages with errors (e^-2 = 0.1363, e^-0.2 = 0.819)

The life of army shoes is normally distributed with mean 8 months and standard deviation 2 months. If 5000 pairs are issued, how many pairs would be expected to need replacement within 12 months.

Suppose that the amount of cosmic radiation to which a person is exposed when flying by jet across USA is a random vertical. having a normal distribution with mean of 4.35m rem and a standard deviation of 0.59m rem. What is the probability that a person will be exposed to more than 5.20 m rem of cosmic radiation of such a flight?

Alpha particles are emitted by a radio active source at an average rate of 5 in a 20 minutes interval. Using Poisson distribution find the probability that there will be atleast 2 emission in a particular 20 minutes interval (e^-5 = 0.0067).

If on an average 1 ship out of 10 do not arrive safely to ports. Find the mean and the standard deviation of ships returning safely out of a total of 500 ships.

10.

A die is thrown 120 times and getting 1 or 5 is considered a success. Find the mean and variance of the number of successes.

11.

Obtain K, μ and σ2 of of the normal distribution whose probability distribution function is f(x) = \(K{ e }^{ -2x^{ 2 }+4x-2 }\), -∞

12.

Find the value of K if X is a normal variate whose p.d.f is given by f(x) = \(\frac { 1 }{ K } \)e^8x-4x2, -∞

13.

If a random variable X follows Poisson distribution such that P(X = 2) = 9. P(X = 4) + 90 P(X = 6) then find the mean and variance.

14.

The standard deviation of a binomial distribution (q +p)¹⁶ is 2. Find its mean.

15.

The probability that an event A happens in one treat of an experiment is 0.4. Three independent treats of the experiment are performed. Find the p!probability that the event A happens at least once.

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Answers Key:

Generic 3

15 x 3 = 45

Let X be the random variable denoting the number of student who passes.
\( \mathrm{p}=\frac{60}{100}=\frac{3}{5}, \mathrm{q}=\frac{40}{100}=\frac{2}{5} \)
\(\mathrm{n}=6, \mathrm{P}(\mathrm{X}=\mathrm{x})=n C_{x} p^{x} q^{n-x} \)
\(\mathrm{P}^{\prime}(\mathrm{X} \geq 4)=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)\)
\( =6 C_{4}\left(\frac{3}{5}\right)^{4}\left(\frac{2}{5}\right)^{2}+6 C_{1}\left(\frac{3}{5}\right)^{4}\left(\frac{2}{5}\right)+\left(\frac{3}{5}\right)^{6} \)
\(=15 \frac{(81 \times 4)}{5^{6}}+\frac{(6)(243)(2)}{5^{6}}+\frac{729}{5^{6}} \)
\( =\frac{8505}{15625}\)

P(X = 1) = 0.4096
P(X = 2) = 0.2048
\( \mathrm{P}(\mathrm{X} =\mathrm{x})=n C_{x} p^{x} q^{n-x} \)
\(5 \mathrm{C}_{1} \mathrm{pq}^{4} =0.4096 \)
\(5 \mathrm{C}_{2} \mathrm{p}^{3} \mathrm{q}^{3} =0.2048 \)
\( \frac{(1)}{(2)} \Rightarrow \frac{5 p q^{4}}{10 p^{2} q^{3}} =\frac{0.4096}{0.2048} \)
\(\frac{q}{2 p} =2 \Rightarrow \mathrm{q}=4 \mathrm{p} \)
\(1-\mathrm{p} =4 \mathrm{p} \)
\(5 \mathrm{p}=1 \Rightarrow \mathrm{p} =\frac{1}{5}=0.2\)

Let X be the random variable denoting the number of heads
\( \mathrm{p}= \frac{1}{2}, \mathrm{q}=\frac{1}{2}, \mathrm{n}=10 \)
\(\mathrm{P}(\mathrm{X} \geq 7) \)
\(= \mathrm{P}(\mathrm{X}=7)+\mathrm{P}(\mathrm{X}=8)+\mathrm{P}(\mathrm{X}=9) +\mathrm{P}(\mathrm{X}=10) \)
\(= 10 C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{3}+10 C_{8}\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{2} +10 C_{9}\left(\frac{1}{2}\right)^{9}\left(\frac{1}{2}\right)^{1}+10 C_{10}\left(\frac{1}{2}\right)^{10} \)
\( =\frac{1}{2^{10}}\left[10 C_{3}+10 C_{2}+10 C_{1}+1\right] \)
\( =\frac{1}{1024}[120+45+10+1] \)
\( =\frac{176}{1024}=\frac{11}{64}\)

Let X be the random variable denoting the number of ships that arrive safely
n = 5
\( \mathrm{p} =\frac{1}{10}, \mathrm{q}=\frac{9}{10} \)
\(\mathrm{P}(\mathrm{X} \geq 4) =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \)
\( =5 \mathrm{C}_{4}\left(\frac{9}{10}\right)^{4}\left(\frac{1}{10}\right)+5 C_{5}\left(\frac{9}{10}\right)^{5} \)
\( =\frac{9^{4}}{10^{5}}[5+9]=0.91854\)

Let X be the random variable denoting the number of pages with errors
n = 400, p = 0.005
\( \lambda=\mathrm{np}=2 \)
\(\mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{e^{-\lambda} \lambda^{x}}{x !} \)
\( \text {(i) } \mathrm{P}(\mathrm{X}=1)=\frac{e^{-2} 2^{1}}{1 !}=0.1363(2) \)
= 0.2726
\(\text {(ii) } \mathrm{P}(\mathrm{X} \leq 3)=\mathrm{P}(\mathrm{X}=0) +\mathrm{P}(\mathrm{X}=1) +\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \)
\(= e^{-2}\left[1+\frac{2}{1 !}+\frac{2^{2}}{2 !}+\frac{2^{3}}{3 !}\right] \)
\(= 0.1363\left(\frac{19}{3}\right)=0.8569 \)

Let X denote the life of army shoes.
Given μ = 8, σ = 2 and N = 5000
When μ = 12, Z = \(\frac { X-\mu }{ \sigma } =\frac { 12-8 }{ 2 } \) = 2
∴ P(X≤12) = P(Z≤2)
= P(-∞ = P(-∞ =0.5+0.4772
P(X≤12) = 0.9772

∴ The probability for a shoe need to be replaces is 0.9772.
∴ Out of 5000 pairs of shoes, number of pairs need to be replaced = 5000 \(\times\) 0.9772
= 4886.

Let X be a random vertical. which is normally distributed
Given that μ = 4.35 and σ = 0.59
When X = 5.20, Z + \(\frac { X-\mu }{ \sigma } =\frac { 5.2-4.35 }{ 0.59 } \)
= \(\frac { 0.85 }{ 0.59 } \) = 1.44

∴ P(X > 5.20) = P(Z > 1.44)
= P(1.44 = P(0 = 0.5 - 0.4251
P(X > 5.20) = 0.749

Average rate of particles emitted in
20 minutes = 5
∴ λ = 5
Let X be the number of particles emitted in 20 minutes
∴ P(X = x) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \), x = 0,1,2,....n
∴ P(X≥2) = 1-P(X<2)
= 1-[P(X = 0) + P(X = 1)]
= 1-\(\left[ \frac { e^{ -5 }.5^{ 0 } }{ 0! } +\frac { e^{ -5 }.5^{ 1 } }{ 1! } \right] \)
= 1-e^-5 (1+5)
= 1-e^-5(6)
= 1-(0.0067)(6)
= 1-0.0402
P(X ≥ 2) = 0.9598

Given n = 500
Letp = Probability of number of ships arrive safely = \(\frac { 9 }{ 10 } \)
∴ q = 1 - p = \(1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } \)
Mean = np = 500 \(\times\) \(\frac { 9 }{ 10 } \)=450
Variance = npq = 450 \(\times\) \(\frac { 1 }{ 10 } \)= 45
∴ Standard deviation =\(\sqrt { npq } =\sqrt { 45 } =3\sqrt { 5 } \).

10.

Given n = 20
P(getting 1 or 5) = \(\frac { 2 }{ 6 } \Rightarrow p=\frac { 1 }{ 3 } \)
∴ q = 1 - p = \(1-\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)
Using Binomial distribution,
Mean = np = 120 \(\times\) \(\frac { 1 }{ 3 } \) = 40
Variance = npq = 40 \(\times\) \(\frac { 2 }{ 3 } =\frac { 80 }{ 3 } \)

11.

Consider -2x²+ 4x - 2 = -2(x²-2x+1)
= -2(x-1)²
∴ \({ e }^{ -2x^{ 2 }+4x-2 }=e^{ -2(x-1)^{ 2 } }\)
\({ e }^{ -\frac { 1 }{ 2 } \frac { (x-1)^{ 2 } }{ \frac { 1 }{ 4 } } }=e^{ -\frac { 1 }{ 2 } \left( \frac { x-1 }{ \frac { 1 }{ 2 } } \right) ^{ 2 } }\)
i.e \(K{ e }^{ -2x^{ 2 }+4x-2 }=\frac { 1 }{ \sigma \sqrt { 2\pi } } .e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) }\)
⇒ \(Ke^{ -\frac { 1 }{ 2 } \left( \frac { x-1 }{ \frac { 1 }{ 2 } } \right) ^{ 2 } }=\frac { 1 }{ \sigma \sqrt { 2\pi } } .e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) }\)
⇒ σ = \(\frac{1}{2}\), μ = 1 and K = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } \)
⇒ K = \(\frac { 1 }{ \frac { 1 }{ 2 } .\sqrt { 2\pi } } \Rightarrow K=\sqrt { \frac { 2 }{ \pi } } \).

12.

The p.d.f for the normal distribution is
f(x) = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } .e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) ^{ 2 } }\), -∞ Given
f(x) = \(\frac { 1 }{ K } { e }^{ 8x-4x^{ 2 } }\)
= \(\frac { 1 }{ K } e^{ -4 }({ x }^{ 2 }-2x)\)
= \(\frac { 1 }{ K } e^{ -4 }({ x }^{ 2 }-2x+1-1)\)
\(\frac { 1 }{ K } e^{ -4 }(x-1)^{ 2 }+4\)
=\(\frac { 1 }{ K } .{ e }^{ 4 }e^{ -\frac { 1 }{ 2 } \frac { x-1^{ 2 } }{ \frac { 1 }{ 8 } } }\)
=\(\frac { 1 }{ 4 } { e }^{ 4 }e^{ -\frac { 1 }{ 2 } \left( \frac { x-1 }{ \frac { 1 }{ \sqrt { 8 } } } \right) ^{ 2 } }\) ...(2)
From (1) & (2), μ = 1, σ = \(\frac { 1 }{ \sqrt { 8 } } \) and
\(\frac { 1 }{ \sigma \sqrt { 2\pi } } =\frac { 1 }{ K } e^{ 4 }\)
\(\frac { 1 }{ \frac { 1 }{ \sqrt { 8 } } .\sqrt { 2 } \pi } =\frac { 1 }{ K } e^{ 4 }\)
⇒ \(\frac { \sqrt { 8 } }{ \sqrt { 2\pi } } =\frac { 1 }{ K } e^{ 4 }\)
\(\sqrt { \frac { 4 }{ \pi } } =\frac { 1 }{ K } { e }^{ 4 }\)
∴ K = \({ e }^{ 4 }\times \sqrt { \frac { \pi }{ 4 } } \)

13.

Given P(X = 2) = 9 P(X = 4) + 90 P(X = 6)
Since X follows Poisson distribution with
P(X,λ) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
\(\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { { e }^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } +\frac { { 90e }^{ -\lambda }.{ \lambda }^{ 6 } }{ 6! } \)

⇒ \(\frac { 1 }{ 2 } =\frac { 3{ \lambda }^{ 2 }+\lambda ^{ 4 } }{ 8 } \)
⇒ λ⁴+3λ² = 4
⇒ λ⁴+ 3λ²-4 = 0
Put λ²= t
⇒ t²+ 3t-4 = 0 ⇒ (t-1)(t + 4) = 0
⇒ t = 1 or t = -4
∴ t = 1 [t = -4 is not possible]
∴ λ²= 1 ⇒ λ = 1
∴ Mean = 1
For Poisson distribution, mean = variance = 1

14.

Given n = 16, S.D = 2 ⇒ \(\sqrt { npq } \) =2
⇒ npq = 4
∴ 16(pq) = 4 ⇒ pq =\(\frac { 4 }{ 16 } =\frac { 1 }{ 4 } \)
⇒ q = \(\frac { 1 }{ 4p } \)...(1)
Since p + q = 1 ⇒ p+\(\frac { 1 }{ 4p } \)=1
⇒ \(\frac { 4{ p }^{ 2 }+1 }{ 4p } \) = 1
⇒ 4p²+1 = 4p ⇒ 4p² -4p+1 = 0
⇒ (2p-1)² = 0 ⇒ 2p-1 = 0
⇒ 2p = 1 ⇒ p = \(\frac { 1 }{ 2 } \)
∴ q = 1-p = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
Mean = np = 16 x \(\frac { 1 }{ 2 } \) = 8

15.

p = 0.4, n = 3
q =1 - P = 1 - 0.4 = 0.6
P(X = x) = nC_x p^xq^n-x
P(X≥1) = P(X = 1) + P(X = 2) + P(X = 3)
=3C₁ (0.4)¹ (0.6)² + 3C₂ (0.4)² (0.6) + 3C₃ (0.4)³ (0.6)⁰
=3 x \(\left( \frac { 4 }{ 10 } \right) \left( \frac { 36 }{ 100 } \right) +3\left( \frac { 16 }{ 100 } \right) \left( \frac { 6 }{ 100 } \right) +\left( \frac { 64 }{ 1000 } \right) \)
= \(\frac { 1 }{ 1000 } \)(432+288+64) =\(\frac { 784 }{ 1000 } \)
P(X≥1) = 0.784