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kolezko [41]
2 years ago
13

last week, Saul ran more than one-fifth the distance that his friend Omar ran. If Saul ran 14 miles last week, how far did Omar

run?
Mathematics
1 answer:
Sonja [21]2 years ago
7 0
1/5 = 14
5/5 = 14 a 5=70
omar ran 70 miles
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A popular resort hotel has 300 rooms and is usually fully booked. About 7% of the time a reservation is canceled before the 6:00
kicyunya [14]

Answer:

8.69% probability that at least 285 rooms will be occupied.

Step-by-step explanation:

For each booked hotel room, there are only two possible outcomes. Either there is a cancelation, or there is not. So we use concepts of the binomial probability distribution to solve this question.

However, we are working with a big sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

A popular resort hotel has 300 rooms and is usually fully booked. This means that n = 300

About 7% of the time a reservation is canceled before the 6:00 p.m. deadline with no pen-alty. What is the probability that at least 285 rooms will be occupied?

Here a success is a reservation not being canceled. There is a 7% probability that a reservation is canceled, and a 100 - 7 = 93% probability that a reservation is not canceled, that is, a room is occupied.  So we use p = 0.93

Approximating the binomial to the normal.

E(X) = np = 300*0.93 = 279

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.93*0.07} = 4.42

The probability that at least 285 rooms will be occupied is 1 subtracted by the pvalue of Z when X = 285. So

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

Z = \frac{285- 279}{4.42}

Z = 1.36

Z = 1.36 has a pvalue of 0.9131.

So there is a 1-0.9131 = 0.0869 = 8.69% probability that at least 285 rooms will be occupied.

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