Find the mean, variance, and standard deviation for each of the values of [tex]$n$[/tex] and [tex]$p$[/tex] when the conditions for the binomial distribution are met. Round your answers to three decimal places as needed.
Part 1 of 4
(a) [tex]$n=118, p=0.74$[/tex]
[tex]\begin{aligned}\text { Mean: } \mu & =87.320 \\ \text { Variance: } \sigma^2 & = \square \\ \text { Standard deviation: } \sigma & =4.766 \end{aligned}$[/tex]
Answer (2)
The variance of the binomial distribution when n = 118 and p = 0.74 is approximately 22.703, and the standard deviation is approximately 4.766. These calculations are based on the formulas for variance and standard deviation in a binomial setting. This information is useful in understanding the variability in the number of successes over repeated trials.
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Calculate the variance using the formula: σ 2 = n p ( 1 − p ) .
Substitute n = 118 and p = 0.74 into the formula: σ 2 = 118 × 0.74 × ( 1 − 0.74 ) .
Calculate the result: σ 2 = 22.7032 .
Round the variance to three decimal places: 22.703 .
Explanation
Problem Analysis We are given a binomial distribution with parameters n = 118 and p = 0.74 . We are asked to find the variance σ 2 .
Variance Formula The formula for the variance of a binomial distribution is given by: σ 2 = n p ( 1 − p ) where n is the number of trials and p is the probability of success in each trial.
Calculation We substitute the given values of n and p into the formula: σ 2 = 118 × 0.74 × ( 1 − 0.74 ) σ 2 = 118 × 0.74 × 0.26 σ 2 = 22.7032 Rounding to three decimal places, we get σ 2 = 22.703 .
Final Answer Therefore, the variance of the binomial distribution is 22.703 .
Examples
Consider a quality control process where you inspect 118 items, and each item has a 74% chance of being defect-free. The variance helps you understand the spread of the number of defective items you might find in each batch of 118. A higher variance indicates a wider range of possible outcomes, which can help in planning and resource allocation.
