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Answer:
Probability that at least 1 of those sampled have access to on-line services at home is 0.9648.
Step-by-step explanation:
We are given that current estimates suggest that 20% of people with home-based computers have access to on-line services.
Suppose that 15 people with home-based computers were randomly and independently sampled.
The above situation can be represented through binomial distribution;
[tex]P(X=r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ;x=0,1,2,3,......[/tex]
where, n = number of trials (samples) taken = 15 people
r = number of success = at least one
p = probability of success which in our question is % of people
who have access to on-line services at home, i.e.; p = 20%
Let X = Number of people who have access to on-line services at home
So, X ~ Binom(n = 15, p = 0.20)
Now, probability that at least 1 of those sampled have access to on-line services at home is given by = P(X [tex]\geq[/tex] 1)
P(X [tex]\geq[/tex] 1) = 1 - P(X = 0)
= [tex]1- \binom{15}{0}\times 0.20^{0} \times (1-0.20)^{15-0}[/tex]
= [tex]1- (1\times 1 \times 0.80^{15})[/tex]
= 1 - 0.0352 = 0.9648
Hence, the required probability is 0.9648.
The following problem can be solved using the binomial distribution.
Binomial distribution
A common discrete distribution is used in statistics, as opposed to a continuous distribution is called a Binomial distribution. It is given by the formula,
[tex]P(x) = ^nC_x p^xq^{(n-x)}[/tex]
Where,
x is the number of successes needed,
n is the number of trials or sample size,
p is the probability of a single success, and
q is the probability of a single failure.
There is 96.481% chance that at least 1 of those sampled have access to online services at home.
Given to us
- The online access computer service industry is growing at an extraordinary rate.
- Current estimates suggest that 20% of people with home-based computers have access to online services.
- Suppose that 15 people with home-based computers were randomly and independently sampled.
Solution
For the values,
- As given, 20% of people with home-based computers have access to online services. therefore, the probability of a person having access to online services is also 20%. thus, p = 20% = 0.2
- We know that for any event the sum of all probability is 1. therefore, the probability of a person not having access to online services is (1-p). thus, q = (1-0.20) = 0.80
- It is already stated in the sentence that the sample is been done on 15 people. therefore, the sample size is n = 15.
- We need the probability that at least 1 of those sampled have access to online services at home, therefore, x ≥ 1.
Probability
Substituting the values in a binomial distribution,
[tex]P(x\geq 1) = 1- P(0)[/tex]
[tex]P(x\geq 1) = 1-\ ^{15}C_0\times [0.20^0]\times [0.80^{(15-10)}]\\\\ P(x\geq 1)=1-\ 0.032\\\\ P(x\geq 1) = 0.96481\\\\ [/tex]
Hence, there is 96.481% chance that at least 1 of those sampled have access to online services at home.
Learn more about Binomial Distribution:
https://brainly.com/question/12734585
