Question

Assume that when adults with smartphones are randomly selected, 5252% use them in meetings or classes. If 66 adult smartphone users are randomly selected, find the probability that at least 22 of them use their smartphones in meetings or classes.

Answer 1

Answer:

The probability is 0.9993

Step-by-step explanation:

This situation follows a Binomial distribution in which we have:

• n identical and independent events: this are the 66 adult smartphone users.
• Two possibles results: success or fail. These are if they use the smartphone in meeting or classes or if they don't.
• Probability p of success: This is the 52% of adults that use them in meeting and classes.

So, the probability that x of the n elements are success is given by:

$P(x)=nCx*p^{x}*(1-p)^{n-x}$

That means that the probability that x adults from the 66 use their smartphone in meeting or classes is:

$P(x)=66Cx*0.52^{x}*(1-0.52)^{66-x}$

Where 66Cx is calculated as:

$66Cx=\frac{66!}{x!(66-x)!}$

Then, the probability that at least 22 of them use their smartphone in meeting or classes is:

P = P(22) + P(23) + P(24) +... + P(64) + P(65) + P(66)

Therefore, replacing x for each number from 22 to 66 on the equation of P(x), and making a sum with all the probabilities, we get that:

P = 0.9993