Question #107139

An electric store has received a shipment of 20 table radios that have connections

for the iPhone. 12 of these have 2 slots and the others have a single slot. Suppose, 6

of the 12 radios are randomly selected to store. Let, X be the number of stored

radios with two slots.

a. What kind of s distribution does X have? (include all the values of the

parameters)

b. Compute, P(X ≤ 3 ).

c. Calculate the mean and standard deviation of X.

d. Calculate the probability that X exceeds its mean value by less than one

standard deviation.

e. Plot the Histogram of X and comment on the shape of the distribution.

Expert's answer

a. X be the number of stored radios with two slots has hypergeometric distribution


P(X=x)=h(x;n,M,N)=(Mx)(N−Mn−x)(Nn)P(X=x)=h(x;n, M, N)={\dbinom{M}{x}\dbinom{N-M}{n-x} \over \dbinom{N}{n}}

Given that N=20,M=12,n=6.N=20, M=12, n=6.


b.

P(X≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)P(X\leq3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)


P(X=0)=(120)(20−126−0)(206)=1(28)38760=2838760P(X=0)={\dbinom{12}{0}\dbinom{20-12}{6-0} \over \dbinom{20}{6}}={1(28) \over 38760}={28 \over 38760}

P(X=1)=(121)(20−126−1)(206)=12(56)38760=67238760P(X=1)={\dbinom{12}{1}\dbinom{20-12}{6-1} \over \dbinom{20}{6}}={12(56) \over 38760}={672 \over 38760}

P(X=2)=(122)(20−126−2)(206)=66(70)38760=462038760P(X=2)={\dbinom{12}{2}\dbinom{20-12}{6-2} \over \dbinom{20}{6}}={66(70) \over 38760}={4620 \over 38760}

P(X=3)=(123)(20−126−3)(206)=220(56)38760=1232038760P(X=3)={\dbinom{12}{3}\dbinom{20-12}{6-3} \over \dbinom{20}{6}}={220(56) \over 38760}={12320\over 38760}

P(X≤3)=2838760+67238760+462038760+1232038760=P(X\leq3)={28 \over 38760}+{672 \over 38760}+{4620 \over 38760}+{12320 \over 38760}=

=1764038760=147323≈0.4551={17640 \over 38760}={147\over 323}\approx0.4551

c. Calculate the mean and standard deviation of X. 


μ=E(X)=n⋅MN=6⋅1220=3.6\mu=E(X)=n\cdot{M \over N}=6\cdot{12 \over 20}=3.6

Var(X)=σ2=(N−nN−1)⋅n⋅MN(1−MN)=Var(X)=\sigma^2=({N-n \over N-1})\cdot n\cdot{M \over N}(1-{M \over N})=

=(20−620−1)⋅6⋅1220(1−1220)=504475=({20-6\over 20-1})\cdot 6\cdot{12 \over 20}(1-{12 \over 20})={504 \over 475}

σ=504475≈1.03\sigma=\sqrt{{504 \over 475}}\approx1.03

d.


μ+σ≈3.6+1.03=4.63\mu+\sigma\approx3.6+1.03=4.63

P(3.6<X≤4.63)=P(X=4)=P(3.6<X\leq4.63)=P(X=4)=

=(124)(20−126−4)(206)=495(28)38760=1386038760=={\dbinom{12}{4}\dbinom{20-12}{6-4} \over \dbinom{20}{6}}={495(28) \over 38760}={13860\over 38760}=

=231646≈0.3576={231\over 646}\approx0.3576

e.



Moderately negative skewed.



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