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A random variable XX has following probability function: \begin{tabular}{|l|...
Aug 22, 2024
Solution by Steps
step 1
To determine the cumulative distribution function (CDF) F(x) F(x) of the random variable X X , we need to calculate the probabilities for X X being less than or equal to each value. The CDF is defined as F(x)=P(X≤x) F(x) = P(X \leq x)
step 2
For x=0 x = 0 : F(0)=P(X≤0)=p(X=0)=0 F(0) = P(X \leq 0) = p(X=0) = 0
step 3
For x=1 x = 1 : F(1)=P(X≤1)=p(X=0)+p(X=1)=0+15=15 F(1) = P(X \leq 1) = p(X=0) + p(X=1) = 0 + \frac{1}{5} = \frac{1}{5}
step 4
For x=2 x = 2 : F(2)=P(X≤2)=p(X=0)+p(X=1)+p(X=2)=0+15+25=35 F(2) = P(X \leq 2) = p(X=0) + p(X=1) + p(X=2) = 0 + \frac{1}{5} + \frac{2}{5} = \frac{3}{5}
step 5
For x=3 x = 3 : F(3)=P(X≤3)=p(X=0)+p(X=1)+p(X=2)+p(X=3)=0+15+25+25=1 F(3) = P(X \leq 3) = p(X=0) + p(X=1) + p(X=2) + p(X=3) = 0 + \frac{1}{5} + \frac{2}{5} + \frac{2}{5} = 1
step 6
Therefore, the cumulative distribution function F(x) F(x) can be summarized as follows: F(x)={0amp;if xlt;015amp;if 0≤xlt;135amp;if 1≤xlt;21amp;if x≥3 F(x) = \begin{cases} 0 & \text{if } x < 0 \\ \frac{1}{5} & \text{if } 0 \leq x < 1 \\ \frac{3}{5} & \text{if } 1 \leq x < 2 \\ 1 & \text{if } x \geq 3 \end{cases}
Answer
The cumulative distribution function F(x) F(x) is defined piecewise as shown above.
Key Concept
The cumulative distribution function (CDF) gives the probability that a random variable is less than or equal to a certain value.
Explanation
The CDF is calculated by summing the probabilities of the random variable up to the specified value, providing a complete view of the distribution.
Solution by Steps
step 1
To find the value of c c , we use the property that the sum of all probabilities must equal 1: P(0)+P(1)+P(2)=1 P(0) + P(1) + P(2) = 1 . Thus, we have: 3c3+(4c−10c2)+(5c−1)=1 3c^3 + (4c - 10c^2) + (5c - 1) = 1
step 2
Simplifying the equation from step 1: 3c3−10c2+9c−2=0 3c^3 - 10c^2 + 9c - 2 = 0 . We can solve this cubic equation for c c
step 3
Using numerical methods or factoring, we find that c=1 c = 1 is a solution. We can verify this by substituting back into the probabilities: P(0)=3(1)3=3 P(0) = 3(1)^3 = 3 , P(1)=4(1)−10(1)2=−6 P(1) = 4(1) - 10(1)^2 = -6 , P(2)=5(1)−1=4 P(2) = 5(1) - 1 = 4 . Since P(1) P(1) must be non-negative, we need to check for other possible values of c c
step 4
After testing possible rational roots, we find c=12 c = \frac{1}{2} satisfies the equation: P(0)=38,P(1)=12,P(2)=38 P(0) = \frac{3}{8}, P(1) = \frac{1}{2}, P(2) = \frac{3}{8} . Thus, c=12 c = \frac{1}{2}
step 5
Now, we calculate the probabilities: P(X < 1) = P(0) = \frac{3}{8}
step 6
For P(1 < X \leq 2) = P(2) = \frac{3}{8}
step 7
For P(0 < X \leq 2) = P(1) + P(2) = \frac{1}{2} + \frac{3}{8} = \frac{7}{8}
Answer
c=12 c = \frac{1}{2} , P(X < 1) = \frac{3}{8} , P(1 < X \leq 2) = \frac{3}{8} , P(0 < X \leq 2) = \frac{7}{8}
Key Concept
The sum of probabilities in a probability mass function must equal 1.
Explanation
We determined c c by ensuring the total probability equals 1 and calculated the required probabilities using the pmf.
Solution by Steps
step 1
To find the probability density function (pdf) of Y Y , we first need to determine the probabilities of Y=1 Y = 1 and Y=−1 Y = -1 based on the values of X X . Since P(X=n)=(12)n P(X = n) = \left( \frac{1}{2} \right)^n , we can calculate the probabilities for even and odd values of X X
step 2
The probability that Y=1 Y = 1 (when X X is even) is given by: P(Y=1)=∑k=1∞P(X=2k)=∑k=1∞(12)2k=1/41−1/4=1/43/4=13. P(Y = 1) = \sum_{k=1}^{\infty} P(X = 2k) = \sum_{k=1}^{\infty} \left( \frac{1}{2} \right)^{2k} = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}.
step 3
The probability that Y=−1 Y = -1 (when X X is odd) is given by: P(Y=−1)=∑k=0∞P(X=2k+1)=∑k=0∞(12)2k+1=1/21−1/4=1/23/4=23. P(Y = -1) = \sum_{k=0}^{\infty} P(X = 2k + 1) = \sum_{k=0}^{\infty} \left( \frac{1}{2} \right)^{2k + 1} = \frac{1/2}{1 - 1/4} = \frac{1/2}{3/4} = \frac{2}{3}.
step 4
Thus, the probability density function of Y Y can be summarized as: P(Y=1)=13,P(Y=−1)=23. P(Y = 1) = \frac{1}{3}, \quad P(Y = -1) = \frac{2}{3}.
Answer
The pdf of Y Y is P(Y=1)=13 P(Y = 1) = \frac{1}{3} and P(Y=−1)=23 P(Y = -1) = \frac{2}{3} .
Key Concept
The relationship between the random variables X X and Y Y allows us to derive the pdf of Y Y based on the probabilities of X X .
Explanation
The pdf of Y Y is derived by summing the probabilities of X X being even and odd, leading to the final probabilities for Y Y .
Solution by Steps
For Q. 11:
step 1
To show that p(x) p(x) is a probability density function (pdf), we need to verify that p(x)≥0 p(x) \geq 0 for all x x and that the integral over its entire range equals 1: ∫−∞∞p(x) dx=1 \int_{-\infty}^{\infty} p(x) \, dx = 1
step 2
Since p(x)=0 p(x) = 0 for x < 0 , we only need to evaluate ∫0∞xe−x2/2 dx \int_{0}^{\infty} x e^{-x^2/2} \, dx . Using the substitution u=x22 u = \frac{x^2}{2} , we find du=x dx du = x \, dx . Thus, the integral becomes ∫0∞e−u du=1 \int_{0}^{\infty} e^{-u} \, du = 1
step 3
Therefore, p(x) p(x) is a valid pdf since it is non-negative and integrates to 1
step 4
The distribution function F(x) F(x) is given by F(x)=∫0xp(t) dt=∫0xte−t2/2 dt F(x) = \int_{0}^{x} p(t) \, dt = \int_{0}^{x} t e^{-t^2/2} \, dt . Using integration by parts, we find F(x)=1−e−x2/2 F(x) = 1 - e^{-x^2/2} for x≥0 x \geq 0 and F(x)=0 F(x) = 0 for x < 0
Answer
p(x) p(x) is a valid pdf and F(x)=1−e−x2/2 F(x) = 1 - e^{-x^2/2} for x≥0 x \geq 0
--- For Q. 12:
step 1
To find the value of a a , we need to ensure that the total area under the pdf equals 1: ∫03f(x) dx=1 \int_{0}^{3} f(x) \, dx = 1 . We calculate the integral piecewise: ∫01ax dx+∫12a dx+∫23(3a−ax) dx \int_{0}^{1} ax \, dx + \int_{1}^{2} a \, dx + \int_{2}^{3} (3a - ax) \, dx
step 2
Evaluating these integrals gives a2+a+(3a−3a2)=1 \frac{a}{2} + a + (3a - \frac{3a}{2}) = 1 , simplifying to 3a2=1 \frac{3a}{2} = 1 , thus a=23 a = \frac{2}{3}
step 3
The cumulative distribution function (CDF) F(x) F(x) is found by integrating f(x) f(x) piecewise: F(x)=∫0xf(t) dt F(x) = \int_{0}^{x} f(t) \, dt . For 0≤x≤1 0 \leq x \leq 1 , F(x)=13x2 F(x) = \frac{1}{3}x^2 ; for 1 < x \leq 2 , F(x)=13+23(x−1)=23x−13 F(x) = \frac{1}{3} + \frac{2}{3}(x - 1) = \frac{2}{3}x - \frac{1}{3} ; for 2 < x \leq 3 , F(x)=23+(3−x)=3−x F(x) = \frac{2}{3} + (3 - x) = 3 - x
step 4
To find the probability that exactly one of x1,x2,x3 x_1, x_2, x_3 is greater than 1.5, we calculate P(X > 1.5) and P(X≤1.5) P(X \leq 1.5) . Using the CDF, P(X > 1.5) = 1 - F(1.5) and P(X≤1.5)=F(1.5) P(X \leq 1.5) = F(1.5) . The probability of exactly one being greater than 1.5 is given by 3 \cdot P(X > 1.5) \cdot (P(X \leq 1.5))^2
Answer
a=23 a = \frac{2}{3} , F(x) F(x) is piecewise defined, and the probability of exactly one of the three observations being greater than 1.5 is calculated using the CDF.
Solution by Steps
For Q. 06:
step 1
To find k k , we use the property that the integral of the pdf over its range must equal 1: ∫0∞kx2e−x dx=1 \int_0^{\infty} kx^2 e^{-x} \, dx = 1
step 2
Evaluating the integral: ∫0∞x2e−x dx=2 \int_0^{\infty} x^2 e^{-x} \, dx = 2 (using the gamma function Γ(n)=(n−1)! \Gamma(n) = (n-1)! ). Thus, k⋅2=1 k \cdot 2 = 1 gives k=12 k = \frac{1}{2}
step 3
The mean E[X] E[X] is calculated as E[X]=∫0∞x⋅kx2e−x dx=k⋅6=3 E[X] = \int_0^{\infty} x \cdot kx^2 e^{-x} \, dx = k \cdot 6 = 3 (since ∫0∞x3e−x dx=6 \int_0^{\infty} x^3 e^{-x} \, dx = 6 )
step 4
The variance Var(X) Var(X) is calculated as Var(X)=E[X2]−(E[X])2 Var(X) = E[X^2] - (E[X])^2 . Here, E[X2]=k⋅24=12 E[X^2] = k \cdot 24 = 12 , so Var(X)=12−9=3 Var(X) = 12 - 9 = 3
Answer
k=12 k = \frac{1}{2} , Mean = 3, Variance = 3
--- For Q. 07:
step 1
To find k k , we use the property that the integral of the pdf over its range must equal 1: ∫01kx(1−x) dx=1 \int_0^1 kx(1-x) \, dx = 1
step 2
Evaluating the integral: ∫01x(1−x) dx=16 \int_0^1 x(1-x) \, dx = \frac{1}{6} . Thus, k⋅16=1 k \cdot \frac{1}{6} = 1 gives k=6 k = 6
step 3
To find b b such that P(X \leq b) = P(X > b) , we set ∫0b6x(1−x) dx=∫b16x(1−x) dx \int_0^b 6x(1-x) \, dx = \int_b^1 6x(1-x) \, dx
step 4
Solving ∫0b6x(1−x) dx=3 \int_0^b 6x(1-x) \, dx = 3 gives b=12 b = \frac{1}{2}
Answer
k=6 k = 6 , b=12 b = \frac{1}{2}
Key Concept
Finding constants in probability density functions and calculating mean and variance.
Explanation
The solutions involve integrating the pdfs to find constants and using properties of expected values to find mean and variance.
Solution by Steps
For Q. 05:
step 1
The number of heads in three flips of a balanced coin follows a binomial distribution with parameters n=3 n = 3 and p=12 p = \frac{1}{2}
step 2
The mean μ \mu is given by μ=n⋅p=3⋅12=32 \mu = n \cdot p = 3 \cdot \frac{1}{2} = \frac{3}{2}
step 3
The variance σ2 \sigma^2 is given by σ2=n⋅p⋅(1−p)=3⋅12⋅12=34 \sigma^2 = n \cdot p \cdot (1 - p) = 3 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}
step 4
The skewness γ1 \gamma_1 is given by γ1=1−2pnp(1−p)=1−13⋅12⋅12=0 \gamma_1 = \frac{1 - 2p}{\sqrt{np(1-p)}} = \frac{1 - 1}{\sqrt{3 \cdot \frac{1}{2} \cdot \frac{1}{2}}} = 0
step 5
The kurtosis γ2 \gamma_2 is given by γ2=(1−6p(1−p))np(1−p)=1−6⋅12⋅123⋅12⋅12=−134=−43 \gamma_2 = \frac{(1 - 6p(1-p))}{np(1-p)} = \frac{1 - 6 \cdot \frac{1}{2} \cdot \frac{1}{2}}{3 \cdot \frac{1}{2} \cdot \frac{1}{2}} = \frac{-1}{\frac{3}{4}} = -\frac{4}{3}
Answer
Mean = 32 \frac{3}{2} , Variance = 34 \frac{3}{4} , Skewness = 0, Kurtosis = −43 -\frac{4}{3}
--- For Q. 06:
step 1
Given n=6 n = 6 and P(X=4)=P(X=2) P(X=4) = P(X=2) , we can express these probabilities using the binomial formula: P(X=k)=(nk)pk(1−p)n−k P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}
step 2
Setting up the equation: (64)p4(1−p)2=(62)p2(1−p)4 \binom{6}{4} p^4 (1-p)^2 = \binom{6}{2} p^2 (1-p)^4
step 3
Simplifying gives 15p4(1−p)2=15p2(1−p)4 15p^4(1-p)^2 = 15p^2(1-p)^4 which reduces to p2(1−p)2=1 p^2(1-p)^2 = 1
step 4
Solving p2(1−p)2=1 p^2(1-p)^2 = 1 leads to p=35 p = \frac{3}{5} or p=25 p = \frac{2}{5}
step 5
The mean μ=n⋅p=6⋅35=185 \mu = n \cdot p = 6 \cdot \frac{3}{5} = \frac{18}{5} , variance σ2=n⋅p(1−p)=6⋅35⋅25=3625 \sigma^2 = n \cdot p(1-p) = 6 \cdot \frac{3}{5} \cdot \frac{2}{5} = \frac{36}{25}
step 6
The skewness γ1=1−2pnp(1−p)=1−656⋅35⋅25=−13625=−56 \gamma_1 = \frac{1 - 2p}{\sqrt{np(1-p)}} = \frac{1 - \frac{6}{5}}{\sqrt{6 \cdot \frac{3}{5} \cdot \frac{2}{5}}} = -\frac{1}{\sqrt{\frac{36}{25}}} = -\frac{5}{6}
step 7
The kurtosis γ2=(1−6p(1−p))np(1−p)=(1−6⋅35⋅25)6⋅35⋅25=−13625=−2536 \gamma_2 = \frac{(1 - 6p(1-p))}{np(1-p)} = \frac{(1 - 6 \cdot \frac{3}{5} \cdot \frac{2}{5})}{6 \cdot \frac{3}{5} \cdot \frac{2}{5}} = \frac{-1}{\frac{36}{25}} = -\frac{25}{36}
Answer
p=35 p = \frac{3}{5} , Mean = 185 \frac{18}{5} , Variance = 3625 \frac{36}{25} , Skewness = −56 -\frac{5}{6} , Kurtosis = −2536 -\frac{25}{36}
--- For Q. 07:
step 1
The probability of getting an even number on a 6-faced die is 36=12 \frac{3}{6} = \frac{1}{2}
step 2
The probability of getting no even numbers in 5 trials is P(X=0)=(1−12)5=(12)5=132 P(X=0) = (1 - \frac{1}{2})^5 = \left(\frac{1}{2}\right)^5 = \frac{1}{32}
step 3
The expected number of sets of 5 trials out of 2500 is 2500⋅P(X=0)=2500⋅132=78.125 2500 \cdot P(X=0) = 2500 \cdot \frac{1}{32} = 78.125
Answer
Expected number of sets = 78.125
--- For Q. 08:
step 1
When two dice are thrown, the total outcomes are 6×6=36 6 \times 6 = 36
step 2
The number of outcomes where the first die exceeds the second die is 1+2+3+4+5=15 1 + 2 + 3 + 4 + 5 = 15
step 3
The probability that the first die exceeds the second die is P=1536=512 P = \frac{15}{36} = \frac{5}{12}
step 4
The average number of times this occurs in 120 trials is 120⋅P=120⋅512=50 120 \cdot P = 120 \cdot \frac{5}{12} = 50
Answer
Average occurrences = 50
Solution by Steps
For Problem 1:
step 1
The variance of a Poisson distribution is equal to its mean, so we have λ=3 \lambda = 3 . To find P(X=2) P(X=2) , we use the formula: P(X=k)=e−λλkk! P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} . Thus, P(X=2)=e−3⋅322! P(X=2) = \frac{e^{-3} \cdot 3^2}{2!}
step 2
Calculating P(X=2) P(X=2) : P(X=2)=e−3⋅92=9e−32≈0.1494 P(X=2) = \frac{e^{-3} \cdot 9}{2} = \frac{9e^{-3}}{2} \approx 0.1494
step 3
To find P(X≥4) P(X \geq 4) , we calculate P(X \geq 4) = 1 - P(X < 4) = 1 - (P(0) + P(1) + P(2) + P(3))
step 4
Calculating P(0),P(1),P(2),P(3) P(0), P(1), P(2), P(3) using the same formula: - P(0)=e−3 P(0) = e^{-3} - P(1)=e−3⋅311!=3e−3 P(1) = \frac{e^{-3} \cdot 3^1}{1!} = 3e^{-3} - P(2)=9e−32 P(2) = \frac{9e^{-3}}{2} (calculated earlier) - P(3)=27e−36=9e−32 P(3) = \frac{27e^{-3}}{6} = \frac{9e^{-3}}{2} Thus, P(X < 4) = e^{-3} + 3e^{-3} + \frac{9e^{-3}}{2} + \frac{9e^{-3}}{2} = 6e^{-3} . Therefore, P(X≥4)=1−6e−3≈0.0498 P(X \geq 4) = 1 - 6e^{-3} \approx 0.0498
Answer
P(X=2)≈0.1494 P(X=2) \approx 0.1494 , P(X≥4)≈0.0498 P(X \geq 4) \approx 0.0498
--- For Problem 2:
step 1
The mean of the Poisson distribution is λ=1.8 \lambda = 1.8 . To find the probability of functioning without a breakdown, we calculate P(X=0)=e−λ=e−1.8 P(X=0) = e^{-\lambda} = e^{-1.8}
step 2
Calculating P(X=0) P(X=0) : P(X=0)≈0.1653 P(X=0) \approx 0.1653
step 3
For one breakdown, P(X=1)=e−λλ11!=1.8e−1.8 P(X=1) = \frac{e^{-\lambda} \lambda^1}{1!} = 1.8e^{-1.8}
step 4
Calculating P(X=1) P(X=1) : P(X=1)≈0.2975 P(X=1) \approx 0.2975
step 5
For at least one breakdown, P(X≥1)=1−P(X=0) P(X \geq 1) = 1 - P(X=0) . Thus, P(X≥1)≈0.8347 P(X \geq 1) \approx 0.8347
Answer
P(X=0)≈0.1653 P(X=0) \approx 0.1653 , P(X=1)≈0.2975 P(X=1) \approx 0.2975 , P(X≥1)≈0.8347 P(X \geq 1) \approx 0.8347
--- For Problem 3:
step 1
The number of defective bulbs follows a Poisson distribution with λ=200⋅0.02=4 \lambda = 200 \cdot 0.02 = 4 . We need to find P(X≤5) P(X \leq 5)
step 2
Using the cumulative distribution function for Poisson, P(X≤k)=∑x=0ke−λλxx! P(X \leq k) = \sum_{x=0}^{k} \frac{e^{-\lambda} \lambda^x}{x!} . Thus, P(X≤5)=e−4∑x=054xx! P(X \leq 5) = e^{-4} \sum_{x=0}^{5} \frac{4^x}{x!}
step 3
Calculating P(X≤5) P(X \leq 5) : P(X≤5)≈0.7845 P(X \leq 5) \approx 0.7845
Answer
P(X≤5)≈0.7845 P(X \leq 5) \approx 0.7845
--- For Problem 4:
step 1
The number of individuals suffering a bad reaction follows a binomial distribution with n=2000 n = 2000 and p=0.001 p = 0.001 . For exactly 3 reactions, we use P(X=3)=(20003)(0.001)3(0.999)1997 P(X=3) = \binom{2000}{3} (0.001)^3 (0.999)^{1997}
step 2
Calculating P(X=3) P(X=3) : P(X=3)≈0.1806 P(X=3) \approx 0.1806
step 3
For more than 2 reactions, P(X > 2) = 1 - (P(X=0) + P(X=1) + P(X=2))
step 4
Calculating P(X=0),P(X=1),P(X=2) P(X=0), P(X=1), P(X=2) using the binomial formula gives P(X > 2) \approx 0.0322
step 5
For none, P(X=0)=(0.999)2000≈0.1353 P(X=0) = (0.999)^{2000} \approx 0.1353
step 6
For more than 1, P(X > 1) = 1 - (P(X=0) + P(X=1)) \approx 0.1806
Answer
P(X=3)≈0.1806 P(X=3) \approx 0.1806 , P(X > 2) \approx 0.0322 , P(X=0)≈0.1353 P(X=0) \approx 0.1353 , P(X > 1) \approx 0.1806
Solution by Steps
step 1
First, we need to calculate the mean μ \mu of the given data using the formula: μ=∑(x⋅f)N \mu = \frac{\sum (x \cdot f)}{N} where N N is the total frequency. Here, N=400 N = 400
step 2
Calculate ∑(x⋅f)=0⋅142+1⋅156+2⋅69+3⋅27+4⋅5+5⋅1=0+156+138+81+20+5=400 \sum (x \cdot f) = 0 \cdot 142 + 1 \cdot 156 + 2 \cdot 69 + 3 \cdot 27 + 4 \cdot 5 + 5 \cdot 1 = 0 + 156 + 138 + 81 + 20 + 5 = 400 . Thus, μ=400400=1 \mu = \frac{400}{400} = 1
step 3
Next, we calculate the variance σ2 \sigma^2 using the formula: σ2=∑(x2⋅f)N−μ2 \sigma^2 = \frac{\sum (x^2 \cdot f)}{N} - \mu^2
step 4
Calculate ∑(x2⋅f)=02⋅142+12⋅156+22⋅69+32⋅27+42⋅5+52⋅1=0+156+276+243+80+25=780 \sum (x^2 \cdot f) = 0^2 \cdot 142 + 1^2 \cdot 156 + 2^2 \cdot 69 + 3^2 \cdot 27 + 4^2 \cdot 5 + 5^2 \cdot 1 = 0 + 156 + 276 + 243 + 80 + 25 = 780 . Thus, σ2=780400−12=1.95−1=0.95 \sigma^2 = \frac{780}{400} - 1^2 = 1.95 - 1 = 0.95
step 5
For skewness γ1 \gamma_1 , we use the formula: γ1=∑(x−μ)3⋅fN⋅σ3 \gamma_1 = \frac{\sum (x - \mu)^3 \cdot f}{N \cdot \sigma^3} First, we need ∑(x−μ)3⋅f \sum (x - \mu)^3 \cdot f
step 6
Calculate ∑(x−1)3⋅f=(0−1)3⋅142+(1−1)3⋅156+(2−1)3⋅69+(3−1)3⋅27+(4−1)3⋅5+(5−1)3⋅1=−142+0+69+8+15+64=14 \sum (x - 1)^3 \cdot f = (0-1)^3 \cdot 142 + (1-1)^3 \cdot 156 + (2-1)^3 \cdot 69 + (3-1)^3 \cdot 27 + (4-1)^3 \cdot 5 + (5-1)^3 \cdot 1 = -142 + 0 + 69 + 8 + 15 + 64 = 14 . Thus, γ1=14400⋅(0.95)3/2≈0.036 \gamma_1 = \frac{14}{400 \cdot (0.95)^{3/2}} \approx 0.036
step 7
For kurtosis γ2 \gamma_2 , we use the formula: γ2=∑(x−μ)4⋅fN⋅σ4−3 \gamma_2 = \frac{\sum (x - \mu)^4 \cdot f}{N \cdot \sigma^4} - 3 First, we need ∑(x−μ)4⋅f \sum (x - \mu)^4 \cdot f
step 8
Calculate ∑(x−1)4⋅f=(0−1)4⋅142+(1−1)4⋅156+(2−1)4⋅69+(3−1)4⋅27+(4−1)4⋅5+(5−1)4⋅1=142+0+69+16+25+16=268 \sum (x - 1)^4 \cdot f = (0-1)^4 \cdot 142 + (1-1)^4 \cdot 156 + (2-1)^4 \cdot 69 + (3-1)^4 \cdot 27 + (4-1)^4 \cdot 5 + (5-1)^4 \cdot 1 = 142 + 0 + 69 + 16 + 25 + 16 = 268 . Thus, γ2=268400⋅(0.95)2−3≈0.1 \gamma_2 = \frac{268}{400 \cdot (0.95)^2} - 3 \approx 0.1
Answer
Mean: 1, Variance: 0.95, Skewness: 0.036, Kurtosis: 0.1
Key Concept
Understanding the properties of a Poisson distribution and calculating its statistical measures.
Explanation
The mean, variance, skewness, and kurtosis were calculated using the frequency data provided, demonstrating the characteristics of the distribution.
Solution by Steps
Q. 01: Probability that the other tube is also good.
step 1
We have a total of 10 tubes: 4 bad and 6 good. We draw 2 tubes, and one is found to be good. We need to find the probability that the other tube is also good
step 2
The possible combinations of drawing 2 tubes are: (G,G),(G,B),(B,G) (G, G), (G, B), (B, G) . Given that one is good, we can ignore the combinations with 2 bad tubes
step 3
The favorable outcomes for the second tube being good are (G,G) (G, G) and (G,B) (G, B) . The probability is calculated as follows: P(other is good∣one is good)=P(G,G)P(G,G)+P(G,B)=6/10⋅5/9(6/10⋅5/9)+(6/10⋅4/9)=59 P(\text{other is good} | \text{one is good}) = \frac{P(G, G)}{P(G, G) + P(G, B)} = \frac{6/10 \cdot 5/9}{(6/10 \cdot 5/9) + (6/10 \cdot 4/9)} = \frac{5}{9}
Answer
59 \frac{5}{9}
--- Q. 02: Are events A, B, and C mutually independent?
step 1
Define the events: A = odd face on the first die, B = odd face on the second die, C = sum is odd. The total outcomes when rolling two dice is 36 36
step 2
Calculate P(A)=P(B)=36=12 P(A) = P(B) = \frac{3}{6} = \frac{1}{2} since there are 3 odd faces on each die
step 3
Calculate P(C) P(C) : The sum is odd if one die is odd and the other is even. Thus, P(C)=1836=12 P(C) = \frac{18}{36} = \frac{1}{2}
step 4
Check independence: P(A∩B)=P(A)⋅P(B)=12⋅12=14 P(A \cap B) = P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} and P(A∩C) P(A \cap C) and P(B∩C) P(B \cap C) must also equal P(A)⋅P(C) P(A) \cdot P(C) and P(B)⋅P(C) P(B) \cdot P(C) respectively
step 5
Since P(A∩B∩C)≠P(A)⋅P(B)⋅P(C) P(A \cap B \cap C) \neq P(A) \cdot P(B) \cdot P(C) , the events are not mutually independent
Answer
No, the events A, B, and C are not mutually independent.
--- Q. 03: Probability that the product is positive.
step 1
We have 6 positive and 8 negative numbers. To have a positive product, we can choose either 0 or an even number of negative numbers
step 2
Calculate the total ways to choose 4 numbers from 14: (144)=1001 \binom{14}{4} = 1001
step 3
Calculate the ways to choose 0 negative numbers: (80)⋅(64)=1⋅15=15 \binom{8}{0} \cdot \binom{6}{4} = 1 \cdot 15 = 15
step 4
Calculate the ways to choose 2 negative numbers: (82)⋅(62)=28⋅15=420 \binom{8}{2} \cdot \binom{6}{2} = 28 \cdot 15 = 420
step 5
Total favorable outcomes = 15+420=435 15 + 420 = 435 . Thus, the probability is: P(positive product)=4351001 P(\text{positive product}) = \frac{435}{1001}
Answer
4351001 \frac{435}{1001}
--- Q. 04: Probability calculations for articles. (i) Both are good.
step 1
Total articles = 10 good + 4 minor defects + 2 major defects = 16. The probability of choosing 2 good articles is: P(both good)=1016⋅915=38 P(\text{both good}) = \frac{10}{16} \cdot \frac{9}{15} = \frac{3}{8}
(ii) Both have major defects.
step 2
The probability of choosing 2 major defect articles is: P(both major defects)=216⋅115=1120 P(\text{both major defects}) = \frac{2}{16} \cdot \frac{1}{15} = \frac{1}{120}
(iii) At least 1 is good.
step 3
The probability of at least 1 good is 1−P(both not good) 1 - P(\text{both not good}) : P(at least 1 good)=1−616⋅515=78 P(\text{at least 1 good}) = 1 - \frac{6}{16} \cdot \frac{5}{15} = \frac{7}{8}
(iv) At most 1 is good.
step 4
The probability of at most 1 good is: P(at most 1 good)=P(0 good)+P(1 good)=616⋅515+(1016⋅615⋅514)=58 P(\text{at most 1 good}) = P(\text{0 good}) + P(\text{1 good}) = \frac{6}{16} \cdot \frac{5}{15} + \left( \frac{10}{16} \cdot \frac{6}{15} \cdot \frac{5}{14} \right) = \frac{5}{8}
(v) Exactly 1 is good.
step 5
The probability of exactly 1 good is: P(exactly 1 good)=1016⋅615=12 P(\text{exactly 1 good}) = \frac{10}{16} \cdot \frac{6}{15} = \frac{1}{2}
(vi) Neither has major defects.
step 6
The probability of neither having major defects is: P(neither major)=1−P(both major)=1−1120=119120 P(\text{neither major}) = 1 - P(\text{both major}) = 1 - \frac{1}{120} = \frac{119}{120}
(vii) Neither is good.
step 7
The probability of neither being good is: P(neither good)=616⋅515=18 P(\text{neither good}) = \frac{6}{16} \cdot \frac{5}{15} = \frac{1}{8}
Answer
(i) 38 \frac{3}{8} , (ii) 1120 \frac{1}{120} , (iii) 78 \frac{7}{8} , (iv) 58 \frac{5}{8} , (v) 12 \frac{1}{2} , (vi) 119120 \frac{119}{120} , (vii) 18 \frac{1}{8}
--- Q. 05: Probability of choosing the false coin.
step 1
We have 3 true coins and 1 false coin. The probability of choosing the false coin is P(F)=14 P(F) = \frac{1}{4}
step 2
If the false coin is chosen, the probability of getting heads 4 times is P(H∣F)=1 P(H|F) = 1 . For true coins, P(H∣T)=(12)4=116 P(H|T) = \left( \frac{1}{2} \right)^4 = \frac{1}{16}
step 3
The total probability of getting heads 4 times is: P(H)=P(H∣F)⋅P(F)+P(H∣T)⋅P(T)=1⋅14+116⋅34=14+364=16+364=1964 P(H) = P(H|F) \cdot P(F) + P(H|T) \cdot P(T) = 1 \cdot \frac{1}{4} + \frac{1}{16} \cdot \frac{3}{4} = \frac{1}{4} + \frac{3}{64} = \frac{16 + 3}{64} = \frac{19}{64}
step 4
Now, we apply Bayes' theorem to find P(F∣H) P(F|H) : P(F∣H)=P(H∣F)⋅P(F)P(H)=1⋅141964=1619 P(F|H) = \frac{P(H|F) \cdot P(F)}{P(H)} = \frac{1 \cdot \frac{1}{4}}{\frac{19}{64}} = \frac{16}{19}
Answer
1619 \frac{16}{19}
--- Q. 06: Probability that all balls are white.
step 1
Let x x be the number of white balls in the bag. The total number of balls is 5. We need to find the probability that all are white given that 2 drawn are white
step 2
The possible scenarios are: (5 white), (4 white, 1 non-white), (3 white, 2 non-white), (2 white, 3 non-white), (1 white, 4 non-white), (0 white)
step 3
The probability of drawing 2 white from 5 white is P(W∣5W)=1 P(W|5W) = 1 . For 4 white, P(W∣4W)=45⋅34=35 P(W|4W) = \frac{4}{5} \cdot \frac{3}{4} = \frac{3}{5} . For 3 white, P(W∣3W)=35⋅24=310 P(W|3W) = \frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}
step 4
The total probability of drawing 2 white is: P(2W)=P(W∣5W)+P(W∣4W)+P(W∣3W)=1+35+310=10+6+310=1910 P(2W) = P(W|5W) + P(W|4W) + P(W|3W) = 1 + \frac{3}{5} + \frac{3}{10} = \frac{10 + 6 + 3}{10} = \frac{19}{10}
step 5
The probability that all are white given that 2 drawn are white is: P(5W∣2W)=P(2W∣5W)⋅P(5W)P(2W)=1⋅161910=1019 P(5W|2W) = \frac{P(2W|5W) \cdot P(5W)}{P(2W)} = \frac{1 \cdot \frac{1}{6}}{\frac{19}{10}} = \frac{10}{19}
Answer
1019 \frac{10}{19}
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