7.4.21-T Queston A certain light arrives on time 83 percent of the time Suppose

Question

7.4.21-T Queston A certain light arrives on time 83 percent of the time Suppose 189 lights are randomly selected Use the normal approximation to the binomial probability that (a) exactly 142 flights are on time (b) at least 142 flights are on time (c) fewer than 141 flights are on time (d) between 141 and 154indasive are on time approi (a) P/142)-00012 (Round to four decimal places as neoded) (b) POX2 142)-09985 (Round to four decimsal places as needed) (c) PCX-141) (Rgund to four decimal places as needed) Enter your answer in the answer box and then click Check Answer

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
mean ( np ) = 189 * 0.83 = 156.87
standard deviation ( npq )= 189*0.83*0.17 = 5.1641
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a.
P(X = 142) = (142-156.87)/5.1641
= -14.87/5.1641= -2.8795
= P ( Z =2.8795) From Standard NOrmal Table
= 0.002
b.
P(X < 142) = (142-156.87)/5.1641
= -14.87/5.1641= -2.8795
= P ( Z <-2.8795) From Standard NOrmal Table
= 0.002
P(X > = 142) = (1 - P(X < 142))
= 1 - 0.002 = 0.998
c.
P(X < 141) = (141-156.87)/5.1641
= -15.87/5.1641= -3.0731
= P ( Z <-3.0731) From Standard NOrmal Table
= 0.0011
d.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 141) = (141-156.87)/5.1641
= -15.87/5.1641 = -3.0731
= P ( Z <-3.0731) From Standard Normal Table
= 0.00106
P(X < 154) = (154-156.87)/5.1641
= -2.87/5.1641 = -0.5558
= P ( Z <-0.5558) From Standard Normal Table
= 0.28919
P(141 < X < 154) = 0.28919-0.00106 = 0.2881

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