A case of wine has 12 bottles, three of which contain spoiled wine. A sample of four bottles is randomly selected from the case.1. Find the probability distribution for X, the number of bottles of spoiled wine in the sample. 2. What are the mean and variance of X

Respuesta :

Answer: (4Cx) × (0.25^x) × (0.75^4-x), mean =1, variance = 0.75

Step-by-step explanation:

Total bottle of wine = 12

Number of spoilt wine = 3

p(spoilt wine) = 3/12 = 1/4 = 0.25, q = 1 - 0.25 = 0.75.

n = number of bottles selected randomly = 4.

Since n = 4, then the experiment is of a binomial.

The probability mass function for a binomial distribution is given as

p(x=r) = nCr × p^r × q^n-r

The distribution for x is given below as

P(X=x) = (4Cx) × (0.25^x) × (0.75^4-x)

The mean for a binomial probability distribution is given as

Mean = np = 4×0.25 = 1

Variance = npq = 4×0.25×0.75 = 0.75

A probability is a number that reflects the chance  that a particular event will occur. Probabilities can be expressed as proportions that range from 0 to 1.

Probability distribution for X is , P(X=3) = 0.012

mean= 1 , variance = 3/4

From question, number of spoiled wine bottle=3

and total number of bottle is 12

So, Probability of success (spoiled wine bottle) p = 3/12 = 1/4

Probability of unsuccessful (non spoiled bottle) q = 1 - 1/4 = 3/4

Since, A sample of four bottles is randomly selected , n = 4

Probability distribution function,  P(X) = ⁿCₓ pⁿ q^(n-x)

Where X represents number of bottles of spoiled wine.

So, P(X=3)= ⁴C₃(1/4)^4(3/4)^1

       P(X=3)= 4(1/4)^4(3/4)

        P(X=3) = 0.012

Mean = np = 4(1/4)=1

Variance = npq = 4 (1/4)(3/4) = 0.75

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