A hospital emergency room has collected a sample of 40 to estimate the mean number of visits per day. The sample standard deviat
1 answer:
<h3>Answer: 8.323 (approximate)</h3>
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Explanation:
s = 32 = sample standard deviation
n = 40 = sample size
Despite not knowing the population standard deviation (sigma), we can still use the Z distribution because n > 30. When n is this large, the student T distribution is approximately the same (more or less) compared to the standard Z distribution. The Z distribution is nicer to work with.
At 90% confidence, the z critical value is roughly z = 1.645. This can be found using either a Z table or a calculator.
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We have these values:
- z = 1.645 (approximate)
- s = 32
- n = 40
Plug these values into the margin of error formula.
The margin of error is roughly 8.323
I rounded to 3 decimal places because z = 1.645 is rounded to 3 decimal places. If your teacher wants some other level of precision, then be sure to follow those instructions.
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Answer:
4.56% of salaries are less than $68,000 or more than $92,000. A percentage lower than 5% is unusual, so it is unusual to find a professor earning less than $68,000 or more than $92,000
Step-by-step explanation:
Problems of normally distributed(bell-shaped) 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.
In this problem, we have that:
What can be said about the percentage of salaries that are less than $68,000 or more than $92,000?
Less than 68,000
pvalue of Z when X = 68000
has a pvalue of 0.0228
2.28% of salaries are less than 68,000.
More than 92,000
1 subtracted by the pvalue of Z when X = 92000
has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% of salaries are more than 92,000.
Less than 68,000 or more than 92,000:
2*2.28 = 4.56% of salaries are less than $68,000 or more than $92,000. A percentage lower than 5% is unusual, so it is unusual to find a professor earning less than $68,000 or more than $92,000