Answer:
The complete explanation and solution is attached below:
Explanation:
An apartment building contains 12 units consisting ot one and two bedroom apartments that rent for $360 and $450 per month, resp
2 answers:
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Answer:
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
That is z with a pvalue of , so Z = 1.645.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 34.09 - 3.25 = 30.84.
The upper end of the interval is the sample mean added to M. So it is 34.09 + 3.25 = 37.34.
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).
Answer:
The minimum value of the bill that is greater than 95% of the bills is $37.87.
Step-by-step explanation:
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.
In this question, we have that:
What are the minimum value of the bill that is greater than 95% of the bills?
This is the 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.
The minimum value of the bill that is greater than 95% of the bills is $37.87.