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 eight of the 20 radios are randomly selected to be stored under a shelf where the radios are displayed, and the remaining that have two slots 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 distribution does X have (name and values of all parameters)?

(b) Compute P(X 2), P(X s 2), and P(X2 2). (Round your answers to four decimal places.)

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

Answer 1

$\small \mathbf{Solution}$

hypergeomertic with N 20, M 10, andn 8

(b)


CN-MC n-r M! formula P(X r) where, C r!(M r)

C 102-10 P(X 2) 20C 8 =

T0C C 6 P(X 2) = 20C

45 210 P(X 2) 125970

$\small P(X=2)=0.075017861$

PX 2)0.0750

P(X2) P(X 0)P(X 1) P(X 2)

$\small P(X\leq 2)=\left [ \frac{_{0}^{10}\textrm{C}*_{8-0}^{20-10}\textrm{C}}{_{8}^{20}\textrm{C}} \right ]+\left [ \frac{_{1}^{10}\textrm{C}*_{8-1}^{20-10}\textrm{C}}{_{8}^{20}\textrm{C}} \right ]+\left [ \frac{_{2}^{10}\textrm{C}*_{8-2}^{20-10}\textrm{C}}{_{8}^{20}\textrm{C}} \right ]$

UCC 10 10 ГС C ОС СТ 20C $8 P(X < 2) 2 + С 8 8


1 45 10120 45 210 P(X 2) 125970 + 125970 125970

ak P(X< 2)I * 45) + (10 120) (45* 210 125970

PX2)0.084901166

$\small P(X\leq 2)\approx \mathbf{{\color{Red} 0.0849}}$

$\small P(X\geq 2)=1-P(X<2)$

$\small P(X\geq 2)=1-\left \{ P(X=0)+P(X=1) \right \}$

C 20-10 20-10 C 8-0 20 8 *8-1 P(X 2) 1 . 2C C 8

$\small P(X\geq 2)=1-\left \{ \left [ \frac{_{0}^{10}\textrm{C}*_{8}^{10}\textrm{C}}{_{8}^{20}\textrm{C}} \right ]+\left [ \frac{_{1}^{10}\textrm{C}*_{7}^{10}\textrm{C}}{_{8}^{20}\textrm{C}} \right ]\right \}$

$\small P(X\geq 2)=1-\left \{ \left [ \frac{1*45}{125970} \right ]+\left [ \frac{10*120}{125970} \right ]\right \}$

$\small P(X\geq 2)=1-\left \{ \left [ \frac{(1*45)+(10*120)}{125970} \right ]\right \}$

$\small P(X\geq 2)=1-\left \{ 0.009883305549\right \}$

$\small P(X\geq 2)=0.990116694$

$\small P(X\geq 2)\approx \mathbf{{\color{Red} 0.9901}}$

$\small P(X\leq 2)$	$\small \mathbf{{\color{Red} 0.0849}}$
$\small P(X\leq 2)$	$\small \mathbf{{\color{Red} 0.0849}}$
P(X 2	$\small \mathbf{{\color{Red} 0.9901}}$

$\small \mathbf{\mathbf{{\color{Blue} }\mathbf{({\color{Blue} c})}}}$

$\small mean\;value=\frac{nM}{N}=\frac{8*10}{20}=\mathbf{{\color{Red} 4}}\;radios$

$\small standard\;deviation=\sqrt{\frac{n*M}{N}*\left ( \frac{N-M}{N} \right )*\left ( \frac{N-n}{N-1} \right )}$

$\small standard\;deviation=\sqrt{\frac{8*10}{20}*\left ( \frac{20-10}{20} \right )*\left ( \frac{20-8}{20-1} \right )}$

$\small standard\;deviation=\sqrt{\frac{80}{20}*\left ( \frac{10}{20} \right )*\left ( \frac{12}{19} \right )}$

$\small standard\;deviation=1.123902974$

$\small standard\;deviation=\mathbf{{\color{Red} 1.12}}\;\mathbf{radios}$