{Exercise 6.31} A Myrtle Beach resort hotel has 120 rooms. In the spring months,
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Question
{Exercise 6.31} A Myrtle Beach resort hotel has 120 rooms. In the spring months, hotel room occupancy is approximately 75%. What is the probability that at least half of the rooms are occupied on a given day (to 4 decimals)? What is the probability that 100 or more rooms are occupied on a given day (to 4 decimals)? What is the probability that 80 or fewer rooms are occupied on a given day (to 4 decimals)?Explanation / Answer
PROBABILITY, NORMAL DISTRIBUTION, LIMITING CASE OF BINOMIAL, n LARGE, 0.1 > p > 0.9, STANDARDIZED VARIABLE z LIMITING CASE OF BINOMIAL: 1. PROBABILITY 60+ ROOMS OCCUPIED (approx.) 100% µ = POPULATION MEAN [n * p] = 120 * 0.75 = 90 n = SAMPLE SIZE [120] p = PROBABILITY OF SUCCESS [0.75] (75%) s = STANDARD DEVIATION [n * p * (1 - p)]^1/2 = [120 * 0.75 * (1 - 0.75)]^1/2 = 4.74 x-bar = NUMBER OF OCCUPIED ROOMS [60] STANDARIZED VARIABLE z = [x-bar - µ]/s = [60 - 90]/4.74 = -6 (APPROX) NORMAL DISTRIBUTION TABLE "LOOK-UP" P(z > -6) = 1.00 (100%) 2. PROBABILITY 100+ ROOMS OCCUPIED (approx.) 1.8% STANDARIZED VARIABLE z = [x-bar - µ]/s = [100 - 90]/4.74 = 2.11 NORMAL DISTRIBUTION TABLE "LOOK-UP" P(z > 2.11) = 0.017507 (1.8%) 3. PROBABILITY 80max ROOMS OCCUPIED (approx) 1.8% STANDARIZED VARIABLE z = [x-bar - µ]/s = [80 - 90]/4.74 = -2.11 NORMAL DISTRIBUTION TABLE "LOOK-UP" P(z < -2.11) = 0.017507 (1.8%)Related Questions
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