Disneyland sells adult passes for $60 and youth passes for $30. In one day, 360 passes were sold for a total of $14,520. How man
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Answer:
99% confidence interval for the population mean is [19.891 , 24.909].
Step-by-step explanation:
We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.
Assuming the population has a normal distribution.
Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;
P.Q. = ~
where, = sample mean age of selected students = 22.4 years
s = sample standard deviation = 3.8 years
n = sample of students = 19
= population mean
<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>
So, 99% confidence interval for the population mean, is ;
P(-2.878 < < 2.878) = 0.99 {As the critical value of t at 18 degree of
freedom are -2.878 & 2.878 with P = 0.5%}
P(-2.878 < < 2.878) = 0.99
P( <
<
) = 0.99
P( <
<
) = 0.99
<u>99% confidence interval for</u> = [
,
]
= [ ,
]
= [19.891 , 24.909]
Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].
Answer:
B.77.4%
Step-by-step explanation:
Mean wage (μ) = $4.50
Standard deviation (σ) = $0.50
For nay given salary X, the z-score is given by:
For X = $3.75, the z-score is:
For X = $5.00, the z-score is:
A z-score of -1.5 corresponds to the 6.68th percentile, while a score of 1 corresponds to the 84.13th percentile. Therefore, the percentage of workers getting paid between $3.75 and $5.00 an hour is:
The answer is alternative B.77.4%