1. 156
2. 15.6
3. 1.56
You just move the decimal point in depending on the number or zeros.
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.
Answer:
The answer is six
Step-by-step explanation:
12 3/5 = 63/5
2 1/10 = 21/10
63/5 divided by 21/10 is equal to 63/5 multiplied by 10/21 which is equal to 630/105 = 6 <-- answer
4/100= x/4,500
18,000/100= $180x2.5 = $450
