Question

An electronics store has received a shipment of 20 table radios that have connections for an iPod or iPhone. Twelve of these have two slots (so they can accommodate both devices), and the other eight have a single slot. Suppose that six 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) What kind of a distribution does X have (name and values of all parameters)?

(b) Compute \( P(X = 2), P(X\leq2), and \hspace{2mm}P(X \geq 2) \).

(c) Calculate the mean value and standard deviation of X.

Answer 1

Solution

X= The number among the radios stored under the display shelf that have two slots :

(a) What kind of a distribution does X have

\( X\sim Hp(n,M,N)\hspace{3mm},n=6,\hspace{2mm}M=12,\hspace{2mm}N=20 \)

Therefore.  \( X\sim Hp(6,12,20) \)

(b) Compute  \( P(X=2),P(X\leq 2),P(X\geq 2) \)

\( \implies P(X=x)=\frac{C_M^x\times C_{N-M}^{n-x}}{C_N^n} \)

\( \bullet \hspace{2mm} P(X=2)=\frac{C_{12}^2\times C_8^4}{C_{20}^6}=\frac{\begin{pmatrix}12\\ 2\end{pmatrix}\times \begin{pmatrix}8\\ 2\end{pmatrix}}{\begin{pmatrix}20\\ 6\end{pmatrix}} \)

Then \( P(X=2)=\frac{\begin{pmatrix}12\\ \:2\end{pmatrix}\times \begin{pmatrix}8\\ \:2\end{pmatrix}}{\begin{pmatrix}20\\ \:6\end{pmatrix}} \)

\( \bullet \hspace{2mm}P(X\leq 2)=P(0)+P(1)+P(2) \)

                      \( =\frac{C_{12}^0\:\times \:C_8^6}{C_{20}^6}+\frac{C_{12}^1\:\times \:\:C_8^5}{C_{20}^6}+\frac{C_{12}^2\:\times \:\:C_8^4}{C_{20}^6} \)

                      \( =\frac{28+12\times \:56+66\times \:70}{C_{20}^6\:} \)

                      \( =\frac{5320}{38760}=0.137 \)

Then. \( P(X\leq 2)=0.137 \)

\( \implies P(X\geq 2)=1-P(X\leq2)+P(X=2) \)

                            \( =1-0.137+0.119=0.98 \)

Therefore.  \( P(X\geq2)=0.98 \)

(c) Calculate the mean value and standard deviation of X.

we have  \( E(X)=n\times \frac{M}{N}=6\times \frac{12}{20}=3.6 \)

\( V(X)=\left(\frac{N-n}{N-1}\right)\times n\:\frac{M}{N}\left(1-\frac{M}{N}\right)=\frac{14}{19}\times 6\times \:\frac{12}{20}\times \:\left(1-\frac{12}{20}\right)=1.02 \)

\( \implies \delta(X)=\sqrt{V(X)}=\sqrt{1.06}=1.02956 \)

Therefore.  \( E(X)=3.6\hspace{2mm},\delta(X)=1.02956 \)

Therefore.

a). \( X\sim Hp(6,12,20) \)

b). \( P(X\geq2)=0.98 \)

c). \( E(X)=3.6\hspace{2mm},\delta(X)=1.02956 \)