you rent an apartment that costs $1600 per month during the first year ,but the rent is set to go up 11.5% per year.what would b
2 answers:
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Answer:
The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 30 - 1 = 29
98% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 29 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.462
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 375 - 41.13 = 333.87 kWh
The upper end of the interval is the sample mean added to M. So it is 375 + 41.13 = 416.13 kWh
The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).
Answer:
The 90th percentile for the distribution of the total contributions is $6,342,525.
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 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.
For sums of size n, the mean is and the standard deviation is
In this question:
The 90th percentile for the distribution of the total contributions
This is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. Then
By the Central Limit Theorem
The 90th percentile for the distribution of the total contributions is $6,342,525.