I guessed “doesn’t matter” but I really want to know why it might be that

Mathematics

1 answer:

chubhunter [2.5K]2 years ago

5 0

I'm sorry, but the spinner that you choose does matter in this case.

For spinner A, there are 5 equal sections with a 1 being in only one of them as well. This gives you a $\frac{1}{5}$ chance of landing on the number 1.

For spinner B, there are 6 equal sections with a 1 being in only one of them. This gives you a $\frac{1}{6}$ chance of landing on the number 1 - a worse chance than if you were to use spinner A.

So for this question, you have a better chance of landing on number 1 if you choose to use spinner A.

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a) A large hotel in Miami has 900 rooms (all rooms are equivalent). During Christmas, the hotel is usually fully booked. However

Olegator [25]

Answer:

14.69% probability that this happens

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$