Question

Chapter 9 4 (10pts) The mean undergraduate cost for tuition, fees, room, and board for four-year institutions was $26,489 for the 2004-2005 academic year. Suppose that a = $3204 and that 36 four-year institutions are randomly selected. Find the probability that the sample mean cost for these 36 schools is less than $25,000 a. b. greater than $26,000 between $24,000 and $26,000 c. 5. (10 pts) Eighty percent of flights arriving in Atlanta for a large US airline are one time. If FAA randomly selects 50 of the airline's flights, find the probability that at least 85% of sampled flights will be on time. а. at most 70% of sampled flights will be on time. b.

Answer 1

4.

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 26489
standard Deviation ( sd )= 3204/ Sqrt ( 36 ) =534
sample size (n) = 36
a.
P(X < 25000) = (25000-26489)/3204/ Sqrt ( 36 )
= -1489/534= -2.7884
= P ( Z <-2.7884) From Standard NOrmal Table
= 0.0026
b.
P(X > 26000) = (26000-26489)/3204/ Sqrt ( 36 )
= -489/534= -0.9157
= P ( Z >-0.9157) From Standard Normal Table
= 0.8201
c.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 24000) = (24000-26489)/3204/ Sqrt ( 36 )
= -2489/534
= -4.661
= P ( Z <-4.661) From Standard Normal Table
= 0
P(X < 26000) = (26000-26489)/3204/ Sqrt ( 36 )
= -489/534 = -0.9157
= P ( Z <-0.9157) From Standard Normal Table
= 0.1799
P(24000 < X < 26000) = 0.1799-0 = 0.1799

5.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.8
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.8*0.2/50)
=0.0566
a.

probability that at least 85%of sampled flights will be on time
P(X < 0.85) = (0.85-0.8)/0.0566
= 0.05/0.0566= 0.8834
= P ( Z <0.8834) From Standard Normal Table
= 0.8115
P(X > = 0.85) = (1 - P(X < 0.85)
= 1 - 0.8115 = 0.1885
b.

probability that at most 70%of sampled flights will be on time
P(X > 0.7) = (0.7-0.8)/0.0566
= -0.1/0.0566 = -1.7668
= P ( Z >-1.767) From Standard Normal Table
= 0.9614
P(X < = 0.7) = (1 - P(X > 0.7)
= 1 - 0.9614 = 0.0386