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: you did not explain it well if you did i would know it but i think i=7 w=18 h=25
Step-by-step explanation:
n+29 is that what you mean
Answer:
<A= 47, <B= 90, <C= 43
Step-by-step explanation:
2x +3 + 4x + 2 + 2x-1 = 180
8x + 4= 180
8x = 176
x =22
2x+3= 44+3= 47
4x+2 = 88+2 = 90
2x-1= 44-1 = 43
