1.United Airlines flight 1832 from chicago to orlando is on time 80% of the time

Question

1.United Airlines flight 1832 from chicago to orlando is on time 80% of the time, according to united states airlines. Suppose 65 flights are randomly selected.A) find the probability that exactly 22 flights are on time.B) Find the probability that at least 2 flights are on time.
2. Assume that the Verbal SAT test score are normally distributed with a mean of 500 and a standard deviation of 100.A) Find the score such that only 10% of the people will score below that score. B)Find the probability that a randomly selected person will score below 650.
3. The university police department must write on average 5 tickets per day to keep department revenues at budgeted levels. Suppose the average number of tickets written per day is known to be 6.4. Find the probability that more than two tickets are written on a randomly selected day from this distribution.

4. For humans, gestation periods are normally distributed with a mean of 266 days and a standard deviation of 16 days. What percentage of humans have a gestational period between 240 and 290 days? B) If a human is told that she is in the 90th percentile at the time of birth how many days was her gestation period?
5. Dutchess county, new york has been experiencing a mean of 0.097 motor vehicle deaths each day. Find the probability that on a given day, there are more than 2 motor vehicle deaths.

Explanation / Answer

1 Given X~Binomial (n=65, p=0.8)
A) find the probability that exactly 22 flights are on time

P(X=22)=0

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B) Find the probability that at least 2 flights are on time.
P(X>=2) = 1-P(X=0)-P(X=1) =1

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2. Given X~Normal(=500 , s=100)

A) Find the score such that only 10% of the people will score below that score

P(X<c)=0.1

--> P((X-)/s <(c-500)/100) =0.1

--> P(Z<(c-500)/100) =0.1

--> (c-500)/100=-1.28 (check standard normal table)

--> c = 500 -1.28*100 =372

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B)Find the probability that a randomly selected person will score below 650.

P(X<650) = P(Z<(650-500)/100)

=P(Z<1.5)

= 0.9332 (check standard normal table)

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3. Find the probability that more than two tickets are written on a randomly selected day from this distribution.

X~Poisson(=6.4)

P(X=x) = (6.4^x)*(exp(-6.4))/x!

So P(X>2) = 1-P(X=0)-P(X=1)-P(X=2) = 0.9536

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4. Given X~Normal(=266,s =16 )

What percentage of humans have a gestational period between 240 and 290 days?

P(240<X<290) = P((240-266)/16 <Z< (290-266)/16)

=P(-1.63<Z<1.5)

= 0.8816 (check standard normal table)

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B) If a human is told that she is in the 90th percentile at the time of birth how many days was her gestation period?

P(X<c)=0.9

--> P(Z<(c-266)/16) =0.9

--> (c-266)/16) =1.28 (check standard normal table)

--> c= 266 + 1.28*16 =286.48

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5.Dutchess county, new york has been experiencing a mean of 0.097 motor vehicle deaths each day. Find the probability that on a given day, there are more than 2 motor vehicle deaths.

Given X~Poisson(=0.097)

P(X>2) = 1-P(X=0) - P(X=1) - P(X=2)= 0.00014

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