Respuesta :
Using the binomial distribution, it is found that:
a) 0.0034 = 0.34% probability that the number expecting a promotion within a month after receiving their degree is six.
b) 0.9958 = 99.58% probability that the number expecting a promotion within a month after receiving their degree is at least seven.
c) 0.0008 = 0.08% probability that the number expecting a promotion within a month after receiving their degree is no more than five.
d) 0.1475 = 14.75% probability that the number expecting a promotion within a month after receiving their degree is between six and nine, inclusive.
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For each student, there are only two possible outcomes, either they expect a promotion, or they do not. The probability of a student expecting a promotion is independent of any other student, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 15 students, thus [tex]n = 15[/tex].
- 75% expect a promotion, thus [tex]p = 0.75[/tex].
Item a:
This is P(X = 6), thus:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{15,6}.(0.75)^{6}.(0.25)^{9} = 0.0034[/tex]
0.0034 = 0.34% probability that the number expecting a promotion within a month after receiving their degree is six.
Item b:
This is:
[tex]P(X \geq 7) = 1 - P(X < 7)[/tex]
In which:
[tex]P(X < 7) = P(X = 6) + P(X = 5) + ... P(X = 0)[/tex]
From item a, we should expect a few of those probabilities to be 0, so:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{15,6}.(0.75)^{6}.(0.25)^{9} = 0.0034[/tex]
[tex]P(X = 5) = C_{15,5}.(0.75)^{5}.(0.25)^{10} = 0.0007[/tex]
[tex]P(X = 4) = C_{15,4}.(0.75)^{4}.(0.25)^{11} = 0.0001[/tex]
The other are 0, thus:
[tex]P(X < 7) = P(X = 6) + P(X = 5) + ... P(X = 0) = 0.034 + 0.0007 + 0.0001 = 0.0042[/tex]
Then
[tex]P(X \geq 7) = 1 - P(X < 7) = 1 - 0.0042 = 0.9958[/tex]
0.9958 = 99.58% probability that the number expecting a promotion within a month after receiving their degree is at least seven.
Item c:
This probability is:
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
Considering item b, the probability of 0 to 3 is 0, thus:
[tex]P(X \leq 5) = P(X = 4) + P(X = 5) = 0.0007 + 0.0001 = 0.0008[/tex]
0.0008 = 0.08% probability that the number expecting a promotion within a month after receiving their degree is no more than five.
Item d:
This probability is:
[tex]P(6 \leq X \leq 9) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{15,6}.(0.75)^{6}.(0.25)^{9} = 0.0034[/tex]
[tex]P(X = 7) = C_{15,7}.(0.75)^{7}.(0.25)^{8} = 0.0131[/tex]
[tex]P(X = 8) = C_{15,8}.(0.75)^{8}.(0.25)^{7} = 0.0393[/tex]
[tex]P(X = 9) = C_{15,9}.(0.75)^{9}.(0.25)^{6} = 0.0917[/tex]
Then
[tex]P(6 \leq X \leq 9) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) = 0.0034 + 0.0131 + 0.0393 + 0.0917 = 0.1475[/tex]
0.1475 = 14.75% probability that the number expecting a promotion within a month after receiving their degree is between six and nine, inclusive.
A similar problem is given at https://brainly.com/question/24863377
