Question

A survey found that three out of ten Americans visited a doctor in any given year. [4a] If 6 people are selected at random, find the probability that no one has visited a doctor last year. [4b] If 6 people are selected at random, find the probability that exactly 1 people has visited a doctor last year. [4c] If 6 people are selected at random, find the probability that at most 1 people have visited a doctor last year. [4d] If 6 people are selected at random, find the probability that at least 2 people have visited a doctor last year. [4e] If 6 people are selected at random, find the mean, variance, standard deviation of the number of people who have visited a doctor last year.

Answer 1

Solution:

Let X be a random variable which represents the number of Americans in a random sample of 6 Americans, who visited a doctor in any given year.

Given that, three out of ten Americans visited a doctor in any given year.

So, probability that an American visited a doctor in any given year = 3/10 = 0.3

Let us consider "An American who visited a doctor in any given year" as success. Hence, we have now only two mutually exclusive outcomes (success and failure).

Probability of success (p) = 0.3

Number of trials (n) = 6

Since, probability of success remains constant in every trial, we have only two mutually exclusive outcomes for every trial, number of trials is finite and outcomes are independent, therefore we can consider that X follows binomial distribution.

According to binomial probability law, the probability of occurrence of exactly x successes in n trials is given by,

Where, p is probability of success.

a) We have to obtain P(X = 0).

We have, n = 6 and p = 0.3

Using binomial probability law we get,

If 6 people are selected at random, the probability that no one has visited a doctor last year is 0.1176.

b) We have to obtain P(X = 1).

We have, n = 6 and p = 0.3

Using binomial probability law we get,

If 6 people are selected at random, the probability that exactly one people has visited a doctor last year is 0.3025.

c) We have to obtain P(X = at most 1).

P(X = at most 1) = P(X ≤ 1)

P(X = at most 1) = P(X = 0) + P(X = 1)

From part (a) we have, P(X = 0) = 0.1176

From part (b) we have, P(X = 1) = 0.3025

Hence, P(X = at most 1) = 0.1176 + 0.3025 = 0.4201

If 6 people are selected at random, the probability that at most 1 people have visited a doctor last year is 0.4201.

d) We have to find P(X = at least 2).

P(X = at least 2) = P(X ≥ 2)

P(X = at least 2) = 1 - P(X < 2) = 1 - [P(X = 0) + P(X = 1)]

From part (a) we have, P(X = 0) = 0.1176

From part (b) we have, P(X = 1) = 0.3025

Hence, P(X = at least 2) = 1 - [0.1176 + 0.3025]

P(X = at least 2) = 1 - 0.4201 = 0.5799

If 6 people are selected at random, the probability that at least 2 people have visited a doctor last year is 0.5798.

e) The mean, variance and standard deviation of binomially distributed random variable X is given as follows:

We have, n = 6 and p = 0.3

Mean = 6 × 0.3 = 1.8

Mean is 1.8.

Variance = 6 × 0.3 × (1 - 0.3) = 1.26

Variance is 1.26.

Standard deviation = √1.26 = 1.1225

The standard deviation is 1.1225.

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