Ikunupe6v

2021-12-18

a) Market research has shown that 60% of persons who are introduced to a certain product actually buy the product. A random sample of 15 persons were introduced to the product.
i. Define the variable of interest for this scenario.
ii. What probability distribution do you think best describes the situation? Why?
iii. Calculate the probability that exactly 9 will buy the product.

eskalopit

Step 1
a) The proportion of persons actually buying the introduced product can be considered as success and proportion of persons not buying the product can be considered as failure.
i. Let X be a variable defined as the number of persons actually buying the introduced product. This is the variable of interest.
ii. The probability distribution suitable for describing this situation is the Binomial Distribution. This is because we are looking for the number of success (number of persons buying the product).
iii. The probability mass function for Binomial distribution is given by:
$P\left(X=x\right){=}^{n}{C}_{x}\cdot {p}^{x}.{\left(1-p\right)}^{n-x}$, where n is the total number of items (in this case persons) and x is the number of success (in this case, persons buying the product).
Step 2
The probability of success is: $p=0.60$.
The sample size is: $n=15$
$P\left(X=9\right){=}^{15}{C}_{9}.{\left(0.60\right)}^{9}.{\left(0.40\right)}^{15-9}$
$=\frac{15!}{9!.6!}.\left(0.010078\right)\left(0.004096\right)$
$=0.2066$

Annie Gonzalez

Step 1
Solution:
1. a.
Here, it is needed to identify the proportion of persons who are introduced to a certain product actually buy the product.
Thus, the variable of interest is the proportion of person who is introduced to a certain product actually buys the product.
Step 2
1. Let X be the number of persons buy the product and n be the sample number of persons were introduced to the product.
From the given information, probability of person who is introduced to a certain product actually buys the product is 0.60 and $n=15$.
Here, persons are independent and probability of success is constant. Hence, X follows binomial distribution with parameters .
Thus, the probability distribution describes the distribution is Binomial.
Step 3
iii. The probability mass function of binomial random variable X is
$P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}\left(1-p{\right)}^{n-x};x=0,1,...,n$
The probability that exactly 9 will buy the product is
$P\left(X=9\right)=\left(\begin{array}{c}15\\ 9\end{array}\right){0.60}^{9}\left(1-0.60{\right)}^{15-9}$
$=0.2066$
Thus, the probability that exactly 9 will buy the product is 0.2066.
Step 4
1. The number of persons who are expected to buy the product if 80 persons are introduced to the product is
$E\left(X\right)=np$
$=80\left(0.60\right)$
$=48$
Thus, the number of persons who are expected to buy the product if 80 persons are introduced to the product is 48.

nick1337

Step 1
NSK
1. A random variable X, is said to follow binomial distribution if the probability mass function of X is,
$P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}\left(1-p{\right)}^{n-x},$ for all PSKx=0,1,...,n, and\ 0<p<1
Here, n is total number of trials and p is the success probability.
The random variable X, that defines the number of persons, who are introduced to a certain product actually buy the product.
Step 3
ii. Here, n= the total number of all persons, who are introduced to the product. There are total 15 persons, who are introduced to the product.
It is obtained that, 60% of persons who are introduced to a certain product actually buy the product. Thus, the success probability is p = 0.6.
In this case, there are two outcomes, such that, whether the person buy the product or not. These two events are independent to each other, and the probabilities of success are independent for each trial.
Thus, X follows Binomial distribution with number of trials 15 and success probability 0.6.
Thus, the binomial probability distribution describe the situation most.
Step 4
iii. The probability that exactly 9 will buy the product is,
$P\left(X=9\right)=\left(\begin{array}{c}15\\ 9\end{array}\right)\left(0.6{\right)}^{9}\left(1-0.6{\right)}^{15-9}$
$=\left(5,005\right)\left(0.010077696\right)\left(0.004096\right)$
$\approx 0.2066$
Thus, the probability that exactly 9 will buy the product is 0.2066.

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