# Assume that when adults with smartphones are randomly selected

Question 1

Assume that when adults with smartphones are randomly selected 62% use them in meetings or classes if 5 adults smartphone users are randomly selected find the probably that exactly 2 of them use their smartphones

Question 2

Assume that when adults with smartphones are randomly selected, 41% use them in meetings or classes. If 9 adult smartphone users are randomly selected, find the probability that exactly 3 of them use their smartphones in meetings or classes.

Question 3

Assume that when adults with smartphones are randomly selected, 58% use them in meetings or classes. If adult smartphone users are randomly selected, find the probability that exactly 7 of them use their smartphones in meetings or classes.

Solution

Given that ,

p = 0.62

1 – p = 1 – 0.62 = 0.38

n = 5

Using binomial probability formula ,

P(X = x) = (x) * px * (1 – p)n – x

P(X = 2) = (2) * (0.62)2 * (0.38)

=  0.21092796

Probability = 0.2109

Binomial Distribution

We know that the data follows the binomial distribution when independent events have two possible possibilities for an outcome to occur (Success or Failure).

In this case, the variable used is discrete. They indicate the following values:

p=0.41n=9p=0.41n=9

The Probability as a function of xx of the Binomial Distribution is given by:

P(X=x)=(nx)⋅px⋅(1−p)n−x,forx=1,2…nP(X=x)=(nx)⋅px⋅(1−p)n−x,forx=1,2…n

The required probability in this case is: P(x=3)=?P(x=3)=?

Using the Binomial Distribution, the probability is:

Atx=3P(X=3)=(93)⋅(0.41)3⋅(1−0.41)9−3(93)=9!3!×(9−3)!=9!3!×6!=9×8×7×6!3!⋅6!=9×8×73!=5046=84P(X=3)=84⋅(0.41)3⋅(0.59)6P(X=3)=84⋅0.068921⋅0.042180534P(X=3)=0.2442Atx=3P(X=3)=(93)⋅(0.41)3⋅(1−0.41)9−3(93)=9!3!×(9−3)!=9!3!×6!=9×8×7×6!3!⋅6!=9×8×73!=5046=84P(X=3)=84⋅(0.41)3⋅(0.59)6P(X=3)=84⋅0.068921⋅0.042180534P(X=3)=0.2442

The formula for probability distribution is given by:

P(x)=nCxpx(1−p)n−xP(x)=nCxpx(1−p)n−x

Where:

• xx is the number of successes,
• pp is the probability of success and,
• nn is the number of trials or the sample size.

In the question, we have the following:

p=0.581−p=1−0.58=0.42p=0.581−p=1−0.58=0.42

Since we are not given the sample size, we will assume that n=8n=8. In this case, the probability that exactly 7 of them (x=7) use their smartphones in meetings or classes is:

P(x=7)=8C7(0.56)7(0.42)8−7P(x=7)=8(0.01727)(0.42)P(x=7)=0.05803

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