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ZanzabumX [31]
3 years ago
12

A hotel manager believes that 14% of the hotel rooms are booked. If the manager is right, what is the probability that the propo

rtion of rooms booked in a sample of 804 rooms would be less than 13%?
Mathematics
1 answer:
lara [203]3 years ago
5 0

Answer:

0.4996

Step-by-step explanation:

We are given;

Population proportion; p = 14% = 0.14

Sample size; n = 804

Sample proportion; p^ = 13% = 0.13

Standard deviation; σ = √np(1 - p)

σ = √(804 × 0.14 × (1 - 0.14))

σ = 9.8388

Formula for z-score is;

z = (p^ - p)/σ

z = (0.13 - 0.14)/9.8388

z = -0.01/9.8388

z = -0.001

From online z-score and probability converter attached, we have;

P(Z < -0.001) = 0.4996

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2 years ago
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Artemon [7]

Answer:

Keenan's z-score was of 0.61.

Rachel's z-score was of 0.81.

Step-by-step explanation:

Z-score:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

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This means that X = 80, \mu = 77, \sigma = 4.9

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Z = \frac{X - \mu}{\sigma}

Z = \frac{80 - 77}{4.9}

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This means that X = 78, \mu = 75, \sigma = 3.7. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{75 - 78}{3.7}

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Rachel's z-score was of 0.81.

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