a) Assume Z N(0,1). Find P(Z < 1.5), P(1.5 < Z < 2.45), and P(|Z| < k) for k = 1

Question

a) Assume Z N(0,1). Find P(Z < 1.5), P(1.5 < Z < 2.45), and P(|Z| < k) for k = 1,2,3.

b) Assume X N( = 70, = 3), then X represents the height in inches of a randomly selected American male.

Find the probability that a randomly selected male is taller than legendary rapper Shaquille O'Neal (7 ft 1 in).

Find the probability that a randomly selected male is between 5 feet and 6 feet tall.

If exactly 25% of American males are taller than Timothy, how tall is Timothy?

c) In atmospheric science, LogNormal distributions are often used to characterize particle size distributions. In one study, the distribution of silicone nanoparticle size (in nm) was found to be approximately lognormal with logmean = 3.91 and logsd = 0.47. Mathematically, Y LogN ( = 3.91, = 0.47). Suppose the desired range of partical sizes is (0.02m, 0.1m), what percentage of silicone nanoparticles do you expect to fall within this range?

Explanation / Answer

Q1.
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 ) = 0
standard Deviation ( sd )= 1
a.
P(X < 1.5) = (1.5-0)/1
= 1.5/1= 1.5
= P ( Z <1.5) From Standard Normal Table
= 0.9332
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -1.5) = (-1.5-0)/1
= -1.5/1 = -1.5
= P ( Z <-1.5) From Standard Normal Table
= 0.0668
P(X < 2.45) = (2.45-0)/1
= 2.45/1 = 2.45
= P ( Z <2.45) From Standard Normal Table
= 0.9929
P(-1.5 < X < 2.45) = 0.9929-0.0668 = 0.926

Q2.
N( = 70, = 3)
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 ) = 70
standard Deviation ( sd )= 3

a.
P(X > 73) = (73-70)/3
= 3/3 = 1
= P ( Z >1) From Standard Normal Table
= 0.1587

b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 60) = (60-70)/3
= -10/3 = -3.3333
= P ( Z <-3.3333) From Standard Normal Table
= 0.0004
P(X < 72) = (72-70)/3
= 2/3 = 0.6667
= P ( Z <0.6667) From Standard Normal Table
= 0.7475
P(60 < X < 72) = 0.7475-0.0004 = 0.7471

c.
P ( Z > x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is 0.6745
P( x-u / (s.d) > x - 70/3) = 0.25
That is, ( x - 70/3) = 0.6745
--> x = 0.6745 * 3+70 = 72.0235

American males are taller than Timothy is 72.0235

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