Respuesta :
Answer:
There is a 5.91% probability that exactly 5 of them will have suffered a hip fracture.
Step-by-step explanation:
For each of these women aged 90, there are only two possible outcomes. Either they will have suffered a hip fracture, or they will not have suffered a hip fracture. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[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.
In this problem we have that:
32 % of all women will fracture their hip by age 90. So [tex]p = 0.32[/tex].
If 8 women aged 90 are selected at random, what is the probability that exactly 5 of them will have suffered a hip fracture?
8 women are selected, so [tex]n = 8[/tex].
The probability is P(X = 5).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{8,5}.(0.32)^{5}.(0.68)^{3} = 0.0591[/tex]
There is a 5.91% probability that exactly 5 of them will have suffered a hip fracture.
Answer: 0.0599
Step-by-step explanation:
Let p denote probability of success, that is probability of choosing someone with fracture by age 90 = 32% = 0.32.
And q denote probability of failure, that is probability of NOT choosing a 90 year old woman with a fractured leg = 68% = 0.68.
Using the probability combination formular for selection, where n is number of samples and r is the number picked among the samples, probability Formula becomes:
= nCr * p^r * q^n-r
Since we're choosing 5women from 8 samples, out Formula becomes
= 8C5 * 0.32^5 * 0.68^3
= 56 * 0.0034 * 0.3144
=0.0599
