Answer:
A=B-5
Step-by-step explanation:
Albert is known as the variable A.
Bill is known as the variable B.
If Albert is 5 years younger than Bill. Then the equation will be
a=b-5
The probability that the mean clock life would differ from the population mean by greater than 12.5 years is 98.30%.
Given mean of 14 years, variance of 25 and sample size is 50.
We have to calculate the probability that the mean clock life would differ from the population mean by greater than 1.5 years.
μ=14,
σ=
=5
n=50
s orσ =5/
=0.7071.
This is 1 subtracted by the p value of z when X=12.5.
So,
z=X-μ/σ
=12.5-14/0.7071
=-2.12
P value=0.0170
1-0.0170=0.9830
=98.30%
Hence the probability that the mean clock life would differ from the population mean by greater than 1.5 years is 98.30%.
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There is a mistake in question and correct question is as under:
What is the probability that the mean clock life would differ from the population mean by greater than 12.5 years?
PQ would be 4 square root 6 times square root of 2, which is also 4 square root 12. Simplified, it would be 8 square root 3
I have a total of 11 pencils = total outcome
a- since we have 6 Red, P(Red)=6/11
b- P(Black) = 3/11
c- P(Green) = 1/11
For the dice:
Total outcome 6 so :
P(3) =1/6
P(1) =1/6
P(less than5) means P( of 1 OR 2 OR 3 Or 4)
Since we have "OR" it means we can add the P(probabilities)====>
P( of 1 OR 2 OR 3 Or 4)=1/6+1/6+1/6+1/6=
= 4/6 = 2/3