Question:

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable X X denote the number of defective items in the sample. If the variance of X X is σ2 \sigma^2 , then 96σ2 96\sigma^2 is equal to _________.

Updated On: Dec 30, 2024
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Correct Answer: 56

Solution and Explanation

Let X X denote the number of defective items in the sample.

xx0123
P(x) P(x) 715 \frac{7}{15} 512 \frac{5}{12} 512 \frac{5}{12} 112 \frac{1}{12}
xi2 x_i^2 0149
P(xi2) P(x_i^2) 0 0 512 \frac{5}{12} 2012 \frac{20}{12} 912 \frac{9}{12}

The mean μ \mu is given by:

μ=P(xi)xi=1812=32 \mu = \sum P(x_i) \cdot x_i = \frac{18}{12} = \frac{3}{2}

Calculate P(xi2) \sum P(x_i^2) :

P(xi2)=3412 \sum P(x_i^2) = \frac{34}{12}

The variance σ2 \sigma^2 is given by:

σ2=P(xi2)μ2 \sigma^2 = \sum P(x_i^2) - \mu^2

σ2=3412(32)2 \sigma^2 = \frac{34}{12} - \left( \frac{3}{2} \right)^2

σ2=341294=34122712=712 \sigma^2 = \frac{34}{12} - \frac{9}{4} = \frac{34}{12} - \frac{27}{12} = \frac{7}{12}

Now, calculate 96σ2 96\sigma^2 :

96σ2=96×712=56 96\sigma^2 = 96 \times \frac{7}{12} = 56

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