A poll was taken of 588 residents in a county. The residents sampled were asked whether they think their local government did a
good job overall. 490 responded ""yes"". Let p denote the proportion of all residents in that county who think their local government did a good job. Construct a 95% confidence interval for p: Round off to two decimal places.
1 answer:
Answer:
[472.29, 507.71]
Step-by-step explanation:
The proportion is
490/588 = 0.8333
The mean can be estimated as
= 490
and the standard deviation can be approximated with
= 9.037
and we just need to find two values a and b such that the area under the Normal curve of mean 490 and standard deviation 9.037 between a and b equals 95% = 0.95
<h3>(See picture)</h3>With the help of a spreadsheet we can easily find these values which are
a = 472.288
b = 507.712
and the interval we are looking for, rounded to 2 decimals, is [472.29, 507.71]
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Answer:
(a) The fraction of employees is 0.84.
(b)
(c)
(d) No. The left part of the distribution would be truncated too much.
Step-by-step explanation:
(a) If the weekly salaries are normally distributed, estimate the fraction of employees that make more than $300 per week.
We have to calculate the z-value and compute the probability
(b) If every employee receives a year-end bonus that adds $100 to the paycheck in the final week, how does this change the normal model for that week?
The mean of the salaries grows $100.
The standard deviation stays the same ($450)
(c) If every employee receives a 5% salary increase for the next year, how does the normal model change?
The increases means a salary X is multiplied by 1.05 (1.05X)
The mean of the salaries grows 5%, to $787.5.
The standard deviation increases by a 5% ($472.5)
(d) If the lowest salary is $300 and the median salary is $525, does a normal model appear appropriate?
No. The left part of the distribution would be truncated too much.