Illene wrapped 5 gifts for her 5 cousins. if she gives them out randomly, what is the probability that every cousin gets the rig
1 answer:
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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
Answer:
a) There is a 10.75% probability of observing less than 60 hits in an hour.
b) The 99th percentile of the distribution of the number of hits is 95.21 hits.
c) There is a 24% probability of observing between 80 and 90 hits an hour
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by
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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.
In this problem, we have that
The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6, so .
a) What’s the probability of observing less than 60 hits in an hour? Use the normal approximation
This is the pvalue of Z when X = 60. So
has a pvalue of 0.1075. This means that there is a 10.75% probability of observing less than 60 hits in an hour.
b) What’s the 99th percentile of the distribution of the number of hits?
What is the value of X when Z has a pvalue of 0.99.
Z = 2.35 has a pvalue of 0.99
So
The 99th percentile of the distribution of the number of hits is 95.21 hits.
c) What’s the probability of observing between 80 and 90 hits an hour?
This is the pvalue of the zscore of X = 90 subtracted by the pvalue of the zscore of X = 80.
For X = 90
has a pvalue of 0.95907
For X = 80
has a pvalue of 0.71904
So
There is a 0.95907 - 0.71904 = 0.24003 = 24% probability of observing between 80 and 90 hits an hour