The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
The lifetime (in hours) of a 60-watt light bulb is a random variable that has a Normal distribution with σ = 30 hours. A random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038 hours.
If in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99% confidence, how many randomly selected 60-watt light bulbs should be tested to achieve this result?
Given Information:
standard deviation = σ = 30 hours
confidence level = 99%
Margin of error = 6 hours
Required Information:
sample size = n = ?
Answer:
sample size = n ≈ 165
Step-by-step explanation:
We know that margin of error is given by
Margin of error = z*(σ/√n)
Where z is the corresponding confidence level score, σ is the standard deviation and n is the sample size
√n = z*σ/Margin of error
squaring both sides
n = (z*σ/Margin of error)²
For 99% confidence level the z-score is 2.576
n = (2.576*30/6)²
n = 164.73
since number of bulbs cannot be in fraction so rounding off yields
n ≈ 165
Therefore, a sample size of 165 bulbs is needed to ensure a margin of error not greater than 6 hours.
Second option. the middle 50%
Answer:
791.28
Step-by-step explanation:
1. 3.14*9²=254.34
2. 3.14*9*19=536.94
3. 536.94+254.34=791.28
Hi,
To find the cost before tax, we must find out how much the tax is. So we multiply 252 x .05 to get 12.6. So we know the tax was 12.6 dollars. Now all we have to do is subtract that from the total to get our answer. 252 - 12.6 = 239.4.
Have a great day!
Steps to solve:
15 + 2a = 35
~Subtract 15 to both sides
2a = 20
~Divide 2 to both sides
a = 10
Best of Luck!
