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tamaranim1 [39]
3 years ago
5

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

eased risk for high blood pressure. (a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker? (b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?
Mathematics
1 answer:
ss7ja [257]3 years ago
3 0

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so \pi = 0.30

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so n = 10

We want to find P(X = 1). So:

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is P(X > 1)

We have that:

P(X \leq 1) + P(X > 1) = 1

P(X > 1) = 1 - P(X \leq 1)

We also have that:

P(X \leq 1) = P(X = 0) + P(X = 1)

P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282

So

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493

Finally

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507

There is a 85.07% probability that more than 1 has the marker.

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