Question

9. Of all customers purchasing automatic garage door openers, 75% purchase chain-driven model. Let X = the number among the next 15 purchasers who select the chain-driven model. a. What is the frequency function (pmf) of X? b. Compute P(X > 11). c. Compute P( 6 ≤ X < 10). d. Compute µ and σ2. e. If the store currently has in stock 10 chain-driven models and 8 shaft-driven models, what is the probability that at least 7 out of the 15 customers select a chain-driven model from this stock?

Answer 1

Answer:

a- $\bf P(X=k)=\binom{15}{k}0.75^k(1-0.75)^{15-k}=\binom{15}{k}0.75^k(0.25)^{15-k}\;(0\leq k\leq 15)$

b- 0.4612

c- 0.1475

d- µ = 11.25 ; $\bf \sigma^2$ = 15*0.75*0.25 = 2.8125

e- 0.3093

Step-by-step explanation:

This is a binomial distribution with probability of “success” (probability of purchasing chain-driven model) is 75%=0.75

a. What is the frequency function (pmf) of X?

$\bf P(X=k)=\binom{15}{k}0.75^k(1-0.75)^{15-k}=\binom{15}{k}0.75^k(0.25)^{15-k}\;(0\leq k\leq 15)$

where  

$\bf \binom{15}{k}$ are combinations of 15 elements taken k at a time

$\bf \binom{15}{k}=\frac{15!}{k!(15-k)!}$

b. Compute P(X > 11)

P(X > 11) = P(X = 12)+P(X = 13)+P(X = 14)+P(X = 15) = 0.4612

c. Compute P( 6 ≤ X < 10).

P( 6 ≤ X < 10) = P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9) = 0.1475

d. Compute µ and $\bf \sigma^2$

µ = 15*0.75 = 11.25

$\bf \sigma^2$ = 15*0.75*0.25 = 2.8125

e. If the store currently has in stock 10 chain-driven models and 8 shaft-driven models, what is the probability that at least 7 out of the 15 customers select a chain-driven model from this stock?

We want

P( 7≤ X ≤ 10) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.3093

Answer 2

PMF is the probability function of discrete distributions. The pmf and the needed probabilities for the given situations are:

• PMF of X = $P(X = x) = \: ^{15}C_x{15}^x(0.25)^{{15}-x}$

• $P(X > 12) \approx 0.461$
• $P(9 \leq X \leq 12) \approx 0.141$
• $\mu = 11.25, \sigma^2 = 2.8125$
• $P(x \geq 7) \approx 0.83$

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability $1- p = q$ (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

The probability that out of n trials, there'd be x successes is given by

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

For the given case, as each purchaser can independently be purchaser of a chain driven model, so if we take:

X = the number among the next 15 purchasers who select the chain-driven model

Then, $X \sim B(n=15, p = 75\% = 0.75)$

where we define success as a purchaser being purchaser of chain driven model, and success is that purchaser not being purchaser of chain driven model.

The pmf of X, thus, is given by:

$P(X =x) = \: ^{15}C_x{15}^x(1-0.75)^{{15}-x}\\\\P(X = x) = \: ^{15}C_x{15}^x(0.25)^{{15}-x}$

Also, we get:

$P(X > 11) = P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)\\\\P(X > 11) =\: ^{15}C_{12}{15}^{12}(0.25)^{3} +\: ^{15}C_{13}{15}^{13}(0.25)^{2} + \: ^{15}C_{14}{15}^{14}(0.25)^{1} + \: ^{15}C_{15}{15}^{15}(0.25)^{0}\\\\P(X > 11) \approx 0.225 + 0.156 + 0.067 + 0.013 = 0.461$

And, similarly,

$P(6 \leq X < 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)\\P(6 \leq X < 10) = \:^{15}C_615}^6(0.25)^{9} + \: ^{15}C_{7}15}^{7}(0.25)^{8} + \\ \: ^{15}C_{8}15}^{8}(0.25)^{7} + \: ^{15}C_{9}15}^{9}(0.25)^{6} \\P(6 \leq X < 10) \approx 0.003 + 0.013 + 0.039 + 0.092\\P(6 \leq X < 10) \approx 0.141$

Mean and variance of binomial distribution are np, and np(1-p) respectively.

For this case, we've n = 15 and p = 0.75, thus:

$\mu = np = 15 \times 0.75 = 11.25\\\sigma^2 = np(1-p) = 11.25 \times 0.25 = 2.8125$

For the case when there are 10 chain-driven models, and 8 shaft-driven models, then the probability of selecting chain driven model (success) would be 10/(10+8) = 10/18 = 5/9 = p, but as there are 15 people still, so n=15, thus, if we take Y = number of people purchasing chain driven model in this case when p = 5/9, then: $Y \sim B(n=15, p=5/9)$

The needed probability is the probability that at least 7 out of the 15 customers select a chain-driven model from this stock.

It can be written symbolically as: $P(X \geq 7)$

We can calculate it as:

$P(X \geq 7) = 1 - P(X < 7) = 1 - ( P(X = 0) + P(X = 1) + ... + P(X = 6))\\P(X \geq 7) = 1 - (\sum_{x=0}^{6} \: ^{15}C_x(5/9)^x(4/9)^{15-x} )\\P(X \geq 7) \approx 1 - 0.17 = 0.83$

Thus, the pmf and the needed probabilities for the given situations are:

• PMF of X = $P(X = x) = \: ^{15}C_x{15}^x(0.25)^{{15}-x}$

• $P(X > 12) \approx 0.461$
• $P(9 \leq X \leq 12) \approx 0.141$
• $\mu = 11.25, \sigma^2 = 2.8125$
• $P(x \geq 7) \approx 0.83$

Learn more about binomial distributions here:

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