Question

Assume that when adults with smartphones are randomly selected, 59% use them in meetings or classes. If 55 adult smartphone users are randomly selected, find the probability that at least 22 of them use their smartphones in meetings or classes. The probability is nothing. (Round to four decimal places as needed.)

Answer #1

P(an adult use smartphone in meetings or classes), p = 0.59

q = 1 - p = 0.41

Sample size, n = 55

P(X < A) = P(Z < (A - mean)/standard deviation)

For the given binomial distribution, Mean = np

= 55x0.59

= 32.45

Standard deviation =

=

= 3.6475

P(at least 22 of 55 use their smartphones in meetings or classes) = P(X 22)

= 1 - P(X < 21.5) (continuity correction of 0.5 is applied)

= 1 - P(Z < (21.5 - 32.45)/3.6475)

= 1 - P(Z < -3.00)

= 1 - 0.0013

= **0.9987**

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use them in meetings or classes. If 10 adult smartphone users are
randomly selected, find the probability that at least 5 of them use
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as needed)
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