Approximately 40% of students enjoy basketball.In a class of 400 students, what is the mean, variance, and standard deviation if we assume normality and use the normal distribution as an approximation of the binomial distribution? Answer choices rounded to the nearest whole number.a.)Mean = 240Variance = 10Standard Deviation = 96b.)Mean = 160Variance = 10Standard Deviation = 96c.)Mean = 160Variance = 96Standard Deviation = 10d.)Mean = 240Variance = 96Standard Deviation = 10
Question
Approximately 40% of students enjoy basketball.In a class of 400 students, what is the mean, variance, and standard deviation if we assume normality and use the normal distribution as an approximation of the binomial distribution? Answer choices rounded to the nearest whole number.a.)Mean = 240Variance = 10Standard Deviation = 96b.)Mean = 160Variance = 10Standard Deviation = 96c.)Mean = 160Variance = 96Standard Deviation = 10d.)Mean = 240Variance = 96Standard Deviation = 10
Solution
The mean (μ), variance (σ²), and standard deviation (σ) for a binomial distribution can be calculated using the following formulas:
Mean (μ) = np Variance (σ²) = np*(1-p) Standard Deviation (σ) = sqrt(σ²)
where: n = number of trials (in this case, the number of students, which is 400) p = probability of success (in this case, the probability that a student enjoys basketball, which is 40% or 0.4)
Let's calculate:
Mean (μ) = np = 4000.4 = 160 Variance (σ²) = np(1-p) = 4000.4(1-0.4) = 96 Standard Deviation (σ) = sqrt(σ²) = sqrt(96) ≈ 10 (rounded to the nearest whole number)
So, the correct answer is:
c.) Mean = 160, Variance = 96, Standard Deviation = 10
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