Answer:
2.28% of faculty members work more than 58.6 hours a week.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The average full-time faculty member in a college works an average of 53 hours per week. Standard deviation of 2.8 hours.
This means that 
What percentage of faculty members work more than 58.6 hours a week?
The proportion is 1 subtracted by the p-value of Z when X = 58.6. So
has a p-value of 0.9772
1 - 0.9772 = 0.0228
0.0228*100% = 2.28%
2.28% of faculty members work more than 58.6 hours a week.
Answer:
$36.50
Step-by-step explanation:
$14.60 divided by 4 = $3.65
$3.65 x 10 = $36.50
Answer:
Atlantic and Atlantic Ocean.
Step-by-step explanation:
They are names, which are proper nouns. All proper nouns are capitilized.
Answer:
(A) With 99% confidence, the proportion that favours the Green initiative is between 418.895 and 423.105 (no approximation; answers were gotten exactly in 3 decimal places).
(B) If many groups of 593 (greater than the initial sample size of 556) randomly selected Americans were surveyed then a different confidence interval would be used or produced from each group.
About 90% of these confidence intervals will contain the true population proportion of Americans who favour the Green initiative and about 10% will not contain the true population proportion.
Step-by-step explanation:
(A) 99% of 421 = 416.79
421 - 416.79 = 4.21
4.21 ÷ 2 = 2.105
(421-2.105), (421+2.105)
The lower and upper limits are:
[418.895 , 423.105]
(B) A wider confidence interval such as 90% will be better suited in a case of multiple samples like this.
