Variance And Standard Deviation

The problem asks for the variance of a random variable , which represents the number of defective items in a random sample of 5 items drawn without replacement from a lot of 12 items containing 3 defectives. After finding the variance in the form of a simplified fraction , we need to compute the value of .

Concept Used:

The random variable follows a hypergeometric distribution. This distribution describes the probability of successes (defective items) in a sample of size drawn without replacement from a population of size containing successes.

The key parameters are:

• : Total population size.
• : Total number of items with the desired feature (successes) in the population.
• : Sample size.

The variance of a random variable following a hypergeometric distribution is given by the formula:

The term is the finite population correction factor.

Step-by-Step Solution:

Step 1: Identify the parameters of the hypergeometric distribution from the problem statement.

• The total number of items in the lot is .
• The number of defective items in the lot is .
• The sample size is .

Step 2: Apply the formula for the variance of the hypergeometric distribution.

The formula for the variance is:

Step 3: Substitute the identified parameter values into the formula.

Now, we simplify each term:

Substitute these simplified fractions back into the variance formula:

Step 4: Calculate the final value of the variance.

Step 5: Identify and and compute .

The variance is given as . We need to check if this fraction is in its simplest form, i.e., .

The prime factorization of the numerator is .

The prime factorization of the denominator is .

Since the numerator and denominator share no common prime factors, . Thus, we have and .

The problem asks for the value of .

The value of is 71.

This problem provides a set of 8 observations, including two unknown values, and . We are given the mean and variance of this dataset and are asked to calculate the value of the expression .

Concept Used:

The solution is based on the definitions of the mean and variance of a dataset.

1. Mean ( ): The mean of a set of observations is the sum of the observations divided by the number of observations.
2. Variance ( ): The variance measures the spread of the data. It is calculated as the mean of the squares of the observations minus the square of the mean.

We will use these formulas to set up two equations with two unknowns, and , and solve for the quantities needed.

Step-by-Step Solution:

Step 1: Use the formula for the mean to find the sum .

The given observations are {6, 4, a, 8, b, 12, 10, 13}.

The number of observations is .

The given mean is .

The sum of all observations is:

Using the mean formula, :

Multiplying both sides by 8:

Solving for :

Step 2: Use the formula for the variance to find the sum of squares .

The given variance is .

The sum of the squares of the observations is:

Using the variance formula, :

Add 81 to both sides:

Multiply both sides by 8:

Solving for :

Step 3: Use the values of and to find the product .

We use the algebraic identity .

Substitute the values we have found:

Final Computation & Result

The problem asks for the value of the expression .

Using the results from our calculations:

Therefore:

The value of is 103.