In a market research survey of 2518 motorists, 252 said that they made an obscene gesture in the previous month.

Mathematics

1 answer:

Lelechka [254]2 years ago

4 0

Answer:

Probability is 89.99%

Step-by-step explanation:

divide 252 by 2518 and you get 10.01% but that represents the people who did make the gesture so instead you would minus 252 from 2518 and that would be 252 - 2518 = 2266 then you would divide 2266 by 2518 and you get 2266/2518 = 0.8999 which is equal to 89.99%

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Suppose monthly rental prices for a one-bedroom apartment in a large city has a distribution that is skewed to the right with a

omeli [17]

Answer:

a) Nothing, beause the distribution of the monthly rental prices are not normal.

b) 1.43% probability that the sample mean rent price will be greater than $900

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

(a) Suppose a one-bedroom rental listing in this large city is selected at random. What can be said about the probability that the listed rent price will be at least $930?

Nothing, beause the distribution of the monthly rental prices are not normal.

(b) Suppose a random sample 30 one-bedroom rental listing in this large city will be selected, the rent price will be recorded for each listing, and the sample mean rent price will be computed. What can be said about the probability that the sample mean rent price will be greater than $900?

Now we can apply the Central Limit Theorem.

$\mu = 880, \sigma = 50, n = 30, s = \frac{50}{\sqrt{30}} = 9.1287$