Answer:
A=5, B=3y
Step-by-step explanation:
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Answer:
189 components must be sampled.
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
That is z with a pvalue of 
, so Z = 2.575.
Now, find the margin of error M as such
Assume that component lifetimes are normally distributed with population standard deviation of 16 hours.
This means that 
How many components must be sampled so that a 99% confidence interval will have margin of error of 3 hours?
n components must be sampled.
n is found when M = 3. So
Rounding up:
189 components must be sampled.
Answer:
0.0918
Step-by-step explanation:
We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The mean and standard deviation of average spending of sample size 25 are
μxbar=μ=95.25
σxbar=σ/√n=27.32/√25=27.32/5=5.464.
So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The z-score associated with average spending $102.5
Z=[Xbar-μxbar]/σxbar
Z=[102.5-95.25]/5.464
Z=7.25/5.464
Z=1.3269=1.33
We have to find P(Xbar>102.5).
P(Xbar>102.5)=P(Z>1.33)
P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)
P(Xbar>102.5)=0.5-0.4082
P(Xbar>102.5)=0.0918.
Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
Answer:
I think the answer is 46 because 9 + 23 + 7 + 7 = 46.
Step-by-step explanation:
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