Respuesta :
Answer:
d) The highest probability occurs when x equals 0.7500
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability, given by the following formula:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]n = 15, p = \frac{1}{20}[/tex]
a) The standard deviation is 0.8441
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15*0.05*0.95} = 0.8441[/tex]
This is correct
b) The number of trials is equal to 15
n is the number of trials and [tex]n = 15[/tex]. So this option is correct.
c) The probability that x equals 1 is 0.3658
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{15,1}.(0.05)^{1}.(0.95)^{14} = 0.3658[/tex]
This option is correct.
d) The highest probability occurs when x equals 0.7500
False. The number of sucesses is a discrete number, that is, 0, 1, 2,...,15. P(X = 0.75), for example, does not exist.
e) The mean equals 0.7500
[tex]E(X) = np = 15*0.05 = 0.75[/tex]
This option is correct.
f) None of the above
d is false
Binomial distribution is distribution tracking number of successes in a set of Bernoulli trials. The true statements for this binomial distribution are:
- The standard deviation of the given distribution is 0.8441
- The number of trials is equal to 15
- The probability that x=1 is 0.3658
- The mean equals 0.7500
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
[tex]X \sim B(n,p)[/tex]
The probability that out of n trials, there'd be x successes is given by
[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]
[tex]E(X) = np\\Var(X) = np(1-p)[/tex]
For the given case, let the random variable be X pertaining the given binomial distribution.
Then, [tex]X \sim B(15, 1/20)[/tex]
- The number of trials = n = 15
Standard deviation = [tex]\sqrt{Var(X)} = \sqrt{np(1-p)} = \sqrt{15 \times 1/20(1-1/20)} \approx \dfrac{16.88}{20} = 0.844[/tex]
- Thus, standard deviation = 0.844 approx
- P(X=1) = [tex]^{15}C_1(1/20)^1(1-1/20)^{14} \approx 0.3685[/tex]
- Highest probability occurs when x = ? . Whatever it be, the value of x must be whole number as it is count of successes which is going to be ranging from 0 to 15 and whole number.
- Mean = E(X) = np = 15 times 1/20 = 0.75
Thus,
The true statements for considered situation's binomial distribution are:
- The standard deviation of the given distribution is 0.8441
- The number of trials is equal to 15
- The probability that x=1 is 0.3658
- The mean equals 0.7500
Learn more about binomial distribution here:
https://brainly.com/question/13609688
