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sasho [114]
2 years ago
12

Rework problem 9 from section 3.2 of your text, involving independent and disjoint events. For this problem, assume that Pr[A∪B]

=0.7 and Pr[A]=0.25 .
Mathematics
1 answer:
tatuchka [14]2 years ago
5 0

Two events are said to be  Disjoint or Mutually Exclusive if the two events can not  happen at the same time.For example when we throw a die getting an even number is disjoint to getting an odd number.

I.e Probability(A∩B)=0

Let me explain this concept through venn diagram.

Pr[A∪B]=0.7, Pr[A]=0.25

Since events are Disjoint

Pr[A∩B]=0

Pr[A∪B]=Pr[A] + Pr[B]

0.7=0.25 +Pr[B]

0.7-0.25=Pr[B]

⇒Pr[B]=0.45=45/100=9/20

Now events are said to be independent if Pr[A and B]=Pr[A] ×Pr[B]

Events are said to be independent if occurrence of one is not affected by occurrence of other.For example getting multiple of 2 as one event and getting multiple of 3 as second event when we throw a die.

Pr[A∪B]=0.7, Pr[A]=0.25

Pr[A∪B]= Pr[A]+ Pr[B]-Pr[A∩B]

But Pr[A∩B]= Pr[A] ×Pr[B]

⇒Pr[A∪B]= Pr[A]+ Pr[B]- Pr[A] ×Pr[B]

⇒0.7=0.25+p-0.25×p

⇒0.7-0.25=p- 0.25 p

⇒0.45=0.75 p

⇒p= 0.45/0.75

⇒p =3/5


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Answer:

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RUDIKE [14]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

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where p and q are the proportion probability of success and q = 1 - p

n is sample size

Then   p = 12 / 30  = 0,4         q =  1 - 0,4    q  =  0,6

And  p*n  =  0,4 * 30  = 12            12 > 10

And  q*n  = 0,6 * 30   = 18            18 > 10

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