Answer:
Step-by-step explanation:
Hello!
Your study variable is X: "number of ColorSmart-5000 that didn't need repairs after 5 years of use, in a sample of 390"
X~Bi (n;ρ)
ρ: population proportion of ColorSmart-5000 that didn't need repairs after 5 years of use. ρ = 0.95
n= 390
x= 303
sample proportion ^ρ: x/n = 303/390 = 0.776 ≅ 0.78
Applying the Central Limit Theorem you approximate the distribution of the sample proportion to normal to obtain the statistic to use.
You are asked to estimate the population proportion of televisions that didn't require repairs with a confidence interval, the formula is:
^ρ±
* √[(^ρ(1-^ρ))/n]
=
= 2.58
0.78±2.58* √[(0.78(1-0.78))/390]
0.0541
[0.726;0.834]
With a confidence level of 99% you'd expect that the interval [0.726;0.834] contains the true value of the proportion of ColorSmart-5000 that didn't need repairs after 5 years of use.
I hope it helps!
Anthony's OT pay will be $118.875 and his overall pay will be $752.875.
Number of hours Anthony has to work in a week = 5 × 8 = 40 hours.
He worked for = 47.50 hours.
Overtime = 47.50 hours - 40 hours = 7.50 hours
Pay per hour = $15.85 / hour
Overtime pay (OT pay) = $15.85 × 7.50 = $118.875
Overall pay = $15.85 × 47.50 = $752.875
Therefore, Anthony's OT pay is $118.875 and his overall pay is $752.875.
Learn more about overtime pay here -
brainly.com/question/19022439
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the answer is 34.83 I believe
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