1. A group of scientist are investigating the airspeed velocity of unladen swall
ID: 3359590 • Letter: 1
Question
1. A group of scientist are investigating the airspeed velocity of unladen swallows. After gathering the data, they have determined that the data is normally distributed with a 10.516 m/s and = 1.455 m/s A) What is the probability that the airspeed velocity of a given unladen swallow is less B) If a given unladen swallow's airspeed velocity is associated with a z-score of 2.20, C) If the probability that a randomly selected swallow has a lower airspeed velocity than D) What is the probability that an unladen swallow will have an airspeed velocity greater than 12 m/s? what is its airspeed velocity? a given swallow is 0.7704, what is the airspeed velocity of the given swallow? than 9.75 m/s?Explanation / Answer
1.
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 ) = 10.516
standard Deviation ( sd )= 1.455
a.probability that the air speed velocity of given unladen swallow is less than 12m/sec
P(X < 12) = (12-10.516)/1.455
= 1.484/1.455= 1.0199
= P ( Z <1.0199) From Standard Normal Table
= 0.8461
b.
Z- score is 2.20
Z = X-u/sd
2.20 = X-10.516/1.455
X= (2.20*1.455)+10.516
X = 13.717 m/sec air speed velocity
c.
P ( Z < x ) = 0.7704
Value of z to the cumulative probability of 0.7704 from normal table is 0.74
P( x-u/s.d < x - 10.516/1.455 ) = 0.7704
That is, ( x - 10.516/1.455 ) = 0.74
--> x = 0.74 * 1.455 + 10.516 = 11.59 m/sec air speed velocity
d.probability that the air speed velocity of given unladen swallow greater than 9.75m/sec
P(X > 9.75) = (9.75-10.516)/1.455
= -0.766/1.455 = -0.5265
= P ( Z >-0.5265) From Standard Normal Table
= 0.7007
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