Question

800 employees of a company are covered under the medical group insurance scheme. Under the term of coverage, 40 employees are identified as belonging to ‘high risk’ category. If 50 employees are selected are random, what is the probability that at the most two are in the high risk category? (a) 0.644 (b) 0.549 (c) 0.744 (d) 0.844

Answer 1

Answer:

b

Step-by-step explanation:

Answer 2

Therefore the probability that at most 2 are in the high-risk category = 0.549 ( Option b )

Given:

Total number of employees in a company who are covered under the medical group insurance scheme = T(n) = 800

Employees who are identified as belonging to 'high risk' category = H(n) = 40

Number of employees selected at random = P(n) = 50

To Find:

The probability that at the most two are in the high-risk category { P ( X ≤ 2 ) }

Solution:

We can simply solve this numerical problem by using the following process.

Number of employees who are not belonging to high risk = 760

q = Probability of number of people who are not at high risk = 760/800 = 0.95

p = Probability of number of people who are at high risk = 40/800 = 0.05

n = Total number of people selected = 50

r = number of people selected

Here n = 50 > 20 which is very large, so Binomial Distribution can't be used rather Poisson distribution can be used.

According to the Poisson Distribution,'

⇒ P ( X = r ) = $e^-^\alpha *\alpha ^r/ r!$

Where α = np = 50 * 0.05 = 2.5

For the probability to be at most 2 high risk category X = 0, X = 1, X = 2 should be found.

For X = 0

⇒  P ( X = 0 ) = $e^-^2^.^5 2.5^0/0!$

⇒  P ( X = 0 ) = 0.082

For X = 1

⇒ P ( X = 1 ) = $e^-^2^.^5 2.5^1 / 1!$

⇒ P ( X = 1 ) = 0.205

For X = 2

⇒ P ( X = 2 ) = $e^-^2^.^5 2.5^2/2!$

⇒ P ( X = 2 ) = 0.2565

Now, P ( X ≤ 2 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2)

⇒ P ( X < 2 ) = 0.082 + 0.205 + 0.2565 = 0.549

Therefore the probability that at most 2 are in the high-risk category = 0.549

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