1.The power consumption of a certain brand light bulb have a normal distribution with mean 60 watts and a standard deviation 2 watts. Find the probability that
(a) 3 bulbs selected randomly each has a power consumption exceeding 61 watts.
X ̅~N(60,4/3), P(X ̅>61)= ..........
(b) 3 bulbs selected randomly have a total power consumption exceeding 183 watts.
2. X~N(μ,9). Find the smallest sample size required to ensure that the probability that X ̅ is within 0.2 of μ is greater than 0.95.
P(-0.2 < X ̅-μ <0.2) > 0.95
P(-0.05√n < Z < 0.05√n) > 0.95......... and solve it?
Please check for me, thank you.
Your message didn't come out clearly but I think the X followed by a box is meant to be X with a bar over it.
For 1 (a) The variance of sampling 3 bulbs isn't a third of the variance of sampling 1 bulb. You are confusing this with the variance of the sample mean which would be a third of the variance of 1 bulb if the central limit theorem applied to sample sizes as small as 3.
To go about this problem you should find the probability of 1 bulb being over 61 then cube that.
(b) here the total power is the sum of 3 normal distributions. The sum of normal distributions is itself a normal distribution with mean equal to the sum of the means and variance equal to the sum of the variances.
2. This is relating to confidence intervals. You know that a confidence interval is where If is to be within 0.2 of then E<0.2
In your answer I'm not sure where the 0.05 comes from.
Did you divide by 3 (which is the standard deviation of our random variable) or by 4? The actual statement should be P(-0.7√n < Z < 0.7√n) > .95. After you check your work, this is simply a matter of finding for which z-value is there 0.025 area to the right. You should be able to take it from there.
Originally Posted by alexander9408