Mathematics, 23.04.2020 01:26 miguel20000000

A group of realtors estimates that 23% of all homes purchased last year were considered investment properties. If a sample of 800 homes sold last year is obtained, what is the probability that at least 175 homes are going to be used as investment property? Round to four decimal places.

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Using the normal approximation to the binomial, it is found that there is a 0.7881 = 78.81% probability that at least 175 homes are going to be used as investment property.

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

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

It 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, which is the percentile of X. The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem:

23% of all homes purchased last year were considered investment properties, hence p = 0.23

Sample of 800 homes, hence n = 800

The mean and the standard deviation for the approximation are given by:

$\mu = np = 800(0.23) = 184$

$\sigma = \sqrt{np(1 - p)} = \sqrt{800(0.23)(0.77)} = 11.9$

Using continuity correction, the probability that at least 175 homes are going to be used as investment property is $P(X \geq 175 - 0.5) = P(X \geq 174.5)$ , which is 1 subtracted by the p-value of Z when X = 174.5.

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

$Z = \frac{174.5 - 184}{11.9}$

Z = -0.8

Z = -0.8 has a p-value of 0.2119.

1 - 0.2119 = 0.7881

0.7881 = 78.81% probability that at least 175 homes are going to be used as investment property.

A similar problem is given at link

A group of realtors estimates that 23% of all homes purchased last year were considered investment p

Answer from: issaaamiaaa15

Probability = 0.77488

Step-by-step explanation:

Since a sample was obtained, this problem is working with the sampling distribution of the sample proportion (p').l which we will need to find the probability.

A sampling distribution of the sample proportion has a mean equal to the population proportion (p).

The standard deviation of the distribution is equal to √((p(1-p))/n).

The sampling distribution in this problem will be normal because the sample size is large.

For this question, the sample proportion(p) is; 23% = 0.23

The sample size(n) is 800, so the standard deviation is;

SD = √((p(1-p))/n) = √((0.23(1-0.23))/800)

SD = 0.0149

The sample proportion is;

(p') = 175/800

p' = 0.21875

Now, for the Z-score;

z = (p' - mean)/SD

z = (0.21875 - 0.23)/ 0.0149

z = - 0.755

Now for the probability, since at least 175,it will be right tail. Thus from the table i attached;

P(z > -0.755) = 1 - 0.22663 = 0.77488

Answer from: Quest

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

Answer from: Quest

Answer might be d.i think

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