QuestionJuly 28, 2025

Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes. If 15 adult smartphone users are randomly selected, find the probability that exactly 11 of them use their smartphones in meetings or classes. The probability is square (Round to four decimal places as needed.)

Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes. If 15 adult smartphone users are randomly selected, find the probability that exactly 11 of them use their smartphones in meetings or classes. The probability is square (Round to four decimal places as needed.)
Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes. If 15 adult smartphone users are randomly selected, find the probability that exactly 11 of them use their smartphones in meetings or classes. The probability is square (Round to four decimal places as needed.)

Solution
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Answer

0.1710 Explanation 1. Identify the distribution This is a binomial distribution problem where n = 15 and p = 0.54. 2. Apply the Binomial Probability Formula Use the formula for binomial probability: **P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}**. Here, k = 11, n = 15, and p = 0.54. 3. Calculate the binomial coefficient \binom{15}{11} = \frac{15!}{11!(15-11)!} = 1365. 4. Compute the probability P(X = 11) = 1365 \times (0.54)^{11} \times (0.46)^4. 5. Perform the calculations P(X = 11) = 1365 \times 0.006366 \times 0.0442 \approx 0.1710.

Explanation

1. Identify the distribution<br /> This is a binomial distribution problem where $n = 15$ and $p = 0.54$.<br /><br />2. Apply the Binomial Probability Formula<br /> Use the formula for binomial probability: **$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$**.<br /> Here, $k = 11$, $n = 15$, and $p = 0.54$.<br /><br />3. Calculate the binomial coefficient<br /> $\binom{15}{11} = \frac{15!}{11!(15-11)!} = 1365$.<br /><br />4. Compute the probability<br /> $P(X = 11) = 1365 \times (0.54)^{11} \times (0.46)^4$.<br /><br />5. Perform the calculations<br /> $P(X = 11) = 1365 \times 0.006366 \times 0.0442 \approx 0.1710$.
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