Question

An electronics store has received a shipment of 20 table radios that have connections for an iPod or iPhone. Ten of these have two slots (so they can accommodate both devices), and the other ten have a single slot. Suppose that seven of the 20 radios are randomly selected to be stored under a shelf where the radios are displayed, and the remaining ones are placed in a storeroom. Let X = the number among the radios stored under the display shelf that have two slots.


(a) Compute
P(X = 2), P(X <= 2), and P(X >= 2).


(b) Calculate the mean value and standard deviation of X. (Round your standard deviation to two decimal places.)

Answer 1

Let \(\mathrm{X}=\) the number among the radios stored under the display shelf that have two slots. The random variable \(X\) has a hyper geometric distribution with the parameters \(n=7, M=10\), and \(N=20\). The probability mass function of hyper geometric distribution can be defined as,

$$ P(X=x)=h(x ; n, M, N) $$

a)

The value of \(P(X=2)\) is,

$$ \begin{aligned} P(X=2) &=h(2 ; 7,10,20) \\ &=\frac{\left(\begin{array}{c} 10 \\ 2 \end{array}\right)\left(\begin{array}{c} 20-10 \\ 7-2 \end{array}\right)}{\left(\begin{array}{c} 20 \\ 7 \end{array}\right)} \\ &=\frac{11340}{77520} \\ &=0.1463 \end{aligned} $$

The value of \(P(X \leq 2)\) is

$$ \begin{aligned} P(X \leq 2) &=P(X=0)+P(X=1)+P(X=2) \\ &=h(0 ; 7,10,20)+h(1 ; 7,10,20)+h(2 ; 7,10,20) \\ &=\frac{\left(\begin{array}{c} 10 \\ 0 \end{array}\right)\left(\begin{array}{c} 20-10 \\ 7-0 \end{array}\right)}{\left(\begin{array}{c} 20 \\ 7 \end{array}\right)}+\frac{\left(\begin{array}{c} 10 \\ 1 \end{array}\right)\left(\begin{array}{c} 20-10 \\ 7-1 \end{array}\right)}{\left(\begin{array}{c} 20 \\ 7 \end{array}\right)}+\frac{\left(\begin{array}{c} 10 \\ 2 \end{array}\right)\left(\begin{array}{c} 20 \\ 7 \end{array}\right)}{\left(\begin{array}{c} 20 \\ 7 \end{array}\right)} \\ &=0.00155+0.02709+0.1463 \\ &=0.1749 \end{aligned} $$

The value of \(P(X \geq 2)\) is

$$ \begin{aligned} P(X \geq 2) &=1-P(X<2) \\ &=1-P(X=0)+P(X=1) \\ &=1-(0.00155+0.02709) \\ &=0.9714 \end{aligned} $$

b)

The mean value of \(X\) is,

$$ \begin{aligned} E(X) &=n \cdot \frac{M}{N} \\ &=7 \cdot \frac{10}{20} \\ &=3.5 \end{aligned} $$

The standard deviation of \(X\) is,

$$ \begin{aligned} S D(X) &=\sqrt{\left(\frac{N-n}{N-1}\right) \cdot n \cdot \frac{M}{N} \cdot\left(1-\frac{M}{N}\right)} \\ &=\sqrt{\left(\frac{20-7}{20-1}\right) \cdot 7 \cdot \frac{10}{20} \cdot\left(1-\frac{10}{20}\right)} \\ &=\sqrt{1.197} \\ &=1.09 \end{aligned} $$