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In Mathematics / College | 2025-07-07

Let N, k and n be the parameters of the hypergeometric experiment. Compute the mean and standard deviation of the hypergeometric random variable X, if N = 180, k = 45, n=17.
The mean of the hypergeometric random variable X is
(Round to the nearest tenth as needed.)
The standard deviation of the hypergeometric random variable X is
(Round to the nearest hundredth as needed.)

Asked by tyram2005

Answer (1)

Calculate the mean using the formula: E [ X ] = n ⋅ ( k / N ) = 17 ⋅ ( 45/180 ) = 4.25 . Round to the nearest tenth: 4.3 .
Calculate the standard deviation using the formula: S D [ X ] = n ⋅ ( k / N ) ⋅ ( 1 − k / N ) ⋅ (( N − n ) / ( N − 1 )) ​ = 17 ⋅ ( 45/180 ) ⋅ ( 1 − 45/180 ) ⋅ (( 180 − 17 ) / ( 180 − 1 )) ​ ≈ 1.703697 . Round to the nearest hundredth: 1.70 .
The mean of the hypergeometric random variable X is 4.3 ​ .
The standard deviation of the hypergeometric random variable X is 1.70 ​ .

Explanation

Understand the problem and provided data We are given a hypergeometric experiment with parameters N = 180 , k = 45 , and n = 17 . We need to compute the mean and standard deviation of the hypergeometric random variable X .

Calculate the mean The mean of a hypergeometric random variable is given by the formula: E [ X ] = n ⋅ N k ​ Substituting the given values, we have: E [ X ] = 17 ⋅ 180 45 ​ = 17 ⋅ 4 1 ​ = 4.25 Rounding to the nearest tenth, we get 4.3 .

Calculate the standard deviation The standard deviation of a hypergeometric random variable is given by the formula: S D [ X ] = n ⋅ N k ​ ⋅ ( 1 − N k ​ ) ⋅ N − 1 N − n ​ ​ Substituting the given values, we have: S D [ X ] = 17 ⋅ 180 45 ​ ⋅ ( 1 − 180 45 ​ ) ⋅ 180 − 1 180 − 17 ​ ​ = 17 ⋅ 4 1 ​ ⋅ 4 3 ​ ⋅ 179 163 ​ ​ S D [ X ] = 4 ⋅ 4 ⋅ 179 17 ⋅ 3 ⋅ 163 ​ ​ = 2864 8313 ​ ​ ≈ 2.90265 ​ ≈ 1.703697 Rounding to the nearest hundredth, we get 1.70 .

State the final answer Therefore, the mean of the hypergeometric random variable X is approximately 4.3 , and the standard deviation is approximately 1.70 .


Examples
Consider a quality control scenario where a batch of 180 items contains 45 defective items. If you randomly select 17 items from the batch, the hypergeometric distribution can help you determine the probability of finding a certain number of defective items in your sample. The mean and standard deviation provide insights into the expected number of defective items and the variability around that expectation, which is crucial for making informed decisions about the batch's overall quality.

Answered by GinnyAnswer | 2025-07-07