Question

According to CTIA, 41% of all U.S. households are wireless-only households (no landline). In a random sample of 20mhouseholds, what is the probability that____________.1. Exactly 5 are wireless-only? 2. Fewer than 3 are wireless-only?3. At least 3 are wireless-only?4. The number of households that are wireless-only is between 5 and 7, inclusive?

Answer 1

Answer:

1) $P(X=5)=(20C5)(0.41)^5 (1-0.41)^{20-5}=0.0656$

2) $P(X<3) = P(X \leq 2)= P(X=0) +P(X=1) +P(X=2)$

And we can find the individual probabilities and we got:

$P(X=0)=(20C0)(0.41)^0 (1-0.41)^{20-0}=0.0000261$

$P(X=1)=(20C1)(0.41)^1 (1-0.41)^{20-1}=0.00036$

$P(X=2)=(20C2)(0.41)^2 (1-0.41)^{20-2}=0.00240$

And adding we got:

$P(X<3) = P(X \leq 2)= P(X=0) +P(X=1) +P(X=2)=0.00279$

3) $P( X \geq 3)$

And using the complment rule we have this:

$P( X \geq 3)= 1-P(X<3) = 1-P(X \leq 2)$

And using the result form the previous result we got:

$P( X \geq 3)= 1-P(X<3) = 1-P(X \leq 2)=1-0.00279 =0.997$

4) $P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)$

And we can find the individual probabilites and we got:

$P(X=5)=(20C2)(0.41)^5 (1-0.41)^{20-5}=0.0656$

$P(X=6)=(20C2)(0.41)^6 (1-0.41)^{20-6}=0.1140$

$P(X=7)=(20C2)(0.41)^7 (1-0.41)^{20-7}=0.1584$

And adding we got:

$P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.338$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.41)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part 1

We want this probability:

$P(X=5)$

We can use th probability mass function and we got:

$P(X=5)=(20C5)(0.41)^5 (1-0.41)^{20-5}=0.0656$

Part 2

We want this probability:

$P(X<3) = P(X \leq 2)= P(X=0) +P(X=1) +P(X=2)$

And we can find the individual probabilities and we got:

$P(X=0)=(20C0)(0.41)^0 (1-0.41)^{20-0}=0.0000261$

$P(X=1)=(20C1)(0.41)^1 (1-0.41)^{20-1}=0.00036$

$P(X=2)=(20C2)(0.41)^2 (1-0.41)^{20-2}=0.00240$

And adding we got:

$P(X<3) = P(X \leq 2)= P(X=0) +P(X=1) +P(X=2)=0.00279$

Part 3

For this case we want this probability:

$P( X \geq 3)$

And using the complment rule we have this:

$P( X \geq 3)= 1-P(X<3) = 1-P(X \leq 2)$

And using the result form the previous result we got:

$P( X \geq 3)= 1-P(X<3) = 1-P(X \leq 2)=1-0.00279 =0.997$

Part 4

For this case we want this probability:

$P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)$

And we can find the individual probabilites and we got:

$P(X=5)=(20C2)(0.41)^5 (1-0.41)^{20-5}=0.0656$

$P(X=6)=(20C2)(0.41)^6 (1-0.41)^{20-6}=0.1140$

$P(X=7)=(20C2)(0.41)^7 (1-0.41)^{20-7}=0.1584$

And adding we got:

$P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.338$

Answer 2

Probability (5 out of 20 use wireless earphones) = 0.656

Probability (less than 3 use wireless earphones) = 0.0027

Probability (more than 3 use wireless earphones) = 0.9973

Probability (households using wireless earphones is between 5 & 7) = 0.9973

Binomial Probability gives probability of r successes out of n trials, with success & failure probability = p & q respectively. Formula =  $N c r. P^r. Q^(n-r)$

• Given : Probability success ie wireless 'p' = 41% = 0.41 , Probability failure 'q' = 59% , Number of trials 'n' = 20 households

1] Prob (r = 5) = $20 c 5. (0.41)^5 . (0.59)^15$  = 0.656

2] Prob (r < 3) = P (r = 1) + P (r = 2) = $20c1 (0.49)^1 (0.51)^19 + 20c2 (0.49)^2 (0.51)^18$ = 0.00036 + 0.00239 = 0.0027

3] Prob (r > 3) = 1 - P (r < 3) = 1 - 0.0027 = 0.9973

4] Prob (5 < r < 7) = P (r = 5) + P (r = 6) + P (r = 7) = $20c5 (0.41)^5 (0.59)^15 + 20c6 (0.41)^6 (0.59)^14 + 20c7 (0.41)^7 (0.59)^13$ = 0.0656 + 0.1140 + 0.1584 = 0.338

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