The seats are divided into 40 sections. If hats are to be given in only 5 sections, what is the probability of a guest sitting i
1 answer:
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Answer:
let b union b be
B={2,4,6,8} and B= {1,2,3,4,5}
then B union B = {1,2,3,4,5,6,8}
Step-by-step explanation:
<em>Venn diagram of b union b is in the attachment </em>
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Answer:
0.0268 = 2.68% probability that the sample average is greater than $229,500
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 and standard deviation
, the zscore of a measure X is given by:
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 and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Single house:
The mean appraised value of a house in this subdivision for all houses is $228,000, with a standard deviation of $8,500, which means that
Sample:
Sample of 120, so, by the central limit theorem,
What is the probability that the sample average is greater than $229,500?
This is 1 subtracted by the pvalue of Z when X = 229.5. So
By the Central Limit Theorem
has a pvalue of 0.9732
1 - 0.9732 = 0.0268
0.0268 = 2.68% probability that the sample average is greater than $229,500