According to reservation records at the hybrid motel/hotel chain Mo'Hotel, 36% of reservations include special requests about room selection, check-in times, or other issues. Let the random variable, x, represent the number of sleeping room reservations that include a special request. The collection is all 2500 sleeping room reservations made last week. #7: When a random sample of 25 such reservations are selected, what is the chance that at least ten (10) but no more than twelve (12) included special requests?

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MrRoyal MrRoyal · Answer 1 · 2022-06-19T14:18:47+02:00

The chance that at least ten (10) but no more than twelve (12) included special requests is 33.59%

The given parameters are:

The chance that at least ten (10) but no more than twelve (12) included special requests is calculated using:

P(10 ≤ x ≤ 12) = P(10) + P(11) + P(12)

Each probability is calculated using:

[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]

So, we have:

[tex]P(10) = ^{25}C_{10} * (36\%)^{10} * (1 - 36\%)^{25-10} =0.14795[/tex]

[tex]P(11) = ^{25}C_{11} * (36\%)^{11} * (1 - 36\%)^{25-11} =0.11348[/tex]

[tex]P(12) = ^{25}C_{12} * (36\%)^{12} * (1 - 36\%)^{25-12} =0.07447[/tex]

So, we have:

P(10 ≤ x ≤ 12) = P(10) + P(11) + P(12)

P(10 ≤ x ≤ 12) = 0.14795 + 0.11348 + 0.07447

Evaluate

P(10 ≤ x ≤ 12) = 0.3359

Express as percentage

P(10 ≤ x ≤ 12) = 33.59%

Hence, the probability is 33.59%