.2IS traffic control system handled an average of 47,529 daily flights during 30
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Question
.2IS traffic control system handled an average of 47,529 daily flights during 30 randomly 14. The U.S. air syster 2010. The historical standard deviation is 6210 flights per day. Based on this selected day information,....OS S1b Find the probability that o eaverage daysnumber of flights will exceed 50,000 ep. ind the probability that the mean of a randomly selected sample of 30 days from 2010 will exceed 50,000. C. 0, 006-172417 30 d. Construct a 99% confidence interval to estimate the average number of/flights per day : , handled by the system. 1- e. Suppose the current system can safely handle 50,000 flights per day. What conclusions can be drawn with these results?Explanation / Answer
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 ) = 47529
standard Deviation ( sd )= 6210
a.
P(X > 50000) = (50000-47529)/6210
= 2471/6210 = 0.3979
= P ( Z >0.3979) From Standard Normal Table
= 0.3453
b.
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 47529
standard Deviation ( sd )= 6210/ Sqrt ( 30 ) =1133.7857
sample size (n) = 30
P(X > 50000) = (50000-47529)/6210/ Sqrt ( 30 )
= 2471/1133.786= 2.1794
= P ( Z >2.1794) From Standard Normal Table
= 0.0147
c.
TRADITIONAL METHOD
given that,
standard deviation, =6210
sample mean, x =47529
population size (n)=30
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 6210/ sqrt ( 30) )
= 1133.79
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 1133.79
= 2222.22
III.
CI = x ± margin of error
confidence interval = [ 47529 ± 2222.22 ]
= [ 45306.78,49751.22 ]
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DIRECT METHOD
given that,
standard deviation, =6210
sample mean, x =47529
population size (n)=30
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 47529 ± Z a/2 ( 6210/ Sqrt ( 30) ) ]
= [ 47529 - 1.96 * (1133.79) , 47529 + 1.96 * (1133.79) ]
= [ 45306.78,49751.22 ]
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interpretations:
1. we are 95% sure that the interval [45306.78 , 49751.22 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
d.
the chaces of probability of such occrances are less than 0.50 and it is considered to be rare event
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