Question

11. Assume that when adults with smartphones are randomly​ selected, 57​% use them in meetings or classes. If 20 adult smartphone users are randomly​ selected, find the probability that exactly 15 of them use their smartphones in meetings or classes.

The probability is _____

​(Round to four decimal places as​ needed.)

12. Assume that when adults with smartphones are randomly​ selected, 58​% use them in meetings or classes. If 10 adult smartphone users are randomly​ selected, find the probability that at least 7 of them use their smartphones in meetings or classes.

The probability is_____

​(Round to four decimal places as​ needed.)

13. A survey showed that 76​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If 8 adults are randomly​ selected, find the probability that no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight​ correction?

The probability that no more than 1 of the 8 adults require eyesight correction is _____.

​(Round to three decimal places as​ needed.)

Answer 1

Please note ⁿC_x = n! / [(n-x)!*x!]

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Also sum of probabilities from 0 till n = 1, i.eP(0) + P(1) + P(2) +.......+P(n) = 1

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(11) Here n = 20, p = 0.57, q = 1 – p = 0.43.

P(X = 15) = ²⁰C₁₅ * (0.57)¹⁵ * (0.43)^20-15 = 0.0497

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(12) Here n = 10, p = 0.58, q = 1 – p = 0.42.

P(At least 7) = P(7) + P(8) + P(9) + P(10)

P(X = 7) = ¹⁰C₇ * (0.58)⁷ * (0.42)^10-7 = 0.1963

P(X = 8) = ¹⁰C₈ * (0.58)⁸ * (0.42)^10-8 = 0.1017

P(X = 9) = ¹⁰C₉ * (0.58)⁹ * (0.42)^10-9 = 0.0312

P(X = 10) = ¹⁰C₁₀ * (0.58)¹⁰ * (0.42)^10-10 = 0.0043

P(At least 7) = 0.1963 + 0.1017 + 0.0312 + 0.0043 = 0.3335

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(13) P(At most 1) = P(0) + P(1)

Here n = 8, p = 0.76, q = 1 – p = 0.24 .

P(X = 0) = ⁸C₀ * (0.76)⁰ * (0.24)^8-0 = 0.000

P(X = 1) = ⁸C₁ * (0.76)¹ * (0.24)^8-1 = 0.000

P(At most 1) = 0.000

Yes, since the value is less than 5%, this is a significantly low number.