2 Suppose that the probability that a flight leaving from Cleveland to Chicago a

Question

2 Suppose that the probability that a flight leaving from Cleveland to Chicago arrives on time 75% of the time. Assume that this flight follows a binomial distribution and that a sample of 18 flights is selected. What is the probability that less than half of the flights out of 18 randomly selected flights will arrive in Chicago on time? Round to the nearest thousandth. 23. Admission to a specialized medical program is determined by an entry exam. The scores of this test are normally distributed with a mean of 500 and a standard deviation of 45. Only students who score in the top 5% will be offered admission. Determine the cutoff scores for the program. Scatter Plot for Average Attendance vs Average Price Average Price 120 Use the regression output below for the NBA's average ticket price and average attendance for the 2016 Season on the remaining problems. Simple linear regression results: Dependent Variable: Average Price Independent Variable: Average Attendance 100 80 Average Price -43.350843+ 0.0055553643 Average Attendance 60 Sample size: 30 R (correlation coefficient) 0.47487197 R-sq = 0.22550338 40 Estimate of error standard deviation: 20.230153 22000 16000 Average 20000 18000 14000

Explanation / Answer

Q22.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 18 * 0.75
= 13.5
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 18 * 0.75 * 0.25
= 3.375
III.
standard deviation = sqrt( variance ) = sqrt(3.375)
=1.837117
a.
P( X < 9) = P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 18 8 ) * 0.75^8 * ( 1- 0.75 ) ^10 + ( 18 7 ) * 0.75^7 * ( 1- 0.75 ) ^11 + ( 18 6 ) * 0.75^6 * ( 1- 0.75 ) ^12 + ( 18 5 ) * 0.75^5 * ( 1- 0.75 ) ^13 + ( 18 4 ) * 0.75^4 * ( 1- 0.75 ) ^14 + ( 18 3 ) * 0.75^3 * ( 1- 0.75 ) ^15 + ( 18 2 ) * 0.75^2 * ( 1- 0.75 ) ^16 + ( 18 1 ) * 0.75^1 * ( 1- 0.75 ) ^17 + ( 18 0 ) * 0.75^0 * ( 1- 0.75 ) ^18
= 0.005422

Q23.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 500
standard Deviation ( sd )= 45
UPPER/TOP
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.644854
P( x-u / (s.d) > x - 500/45) = 0.05
That is, ( x - 500/45) = 1.644854
--> x = 1.644854 * 45+500 = 574.018413

Q24.
correlation = 0.47.
using this it is partially positively correlated.

Q25.
Y' = -43.350843 + .0055553643 * average attendence

Q26.
slope = b1 = 0.0055553643

Q27.
intercept = bo = -43.350843

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