1) According to a study done by De Anza students, the height for Asian adult mal
ID: 2922019 • Letter: 1
Question
1) According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.
a) Give the distribution of X.
b) Find the probability that the person is between 64 and 68 inches.
Write the probability statement.
P(____< x <_____)
What is the probability? (Round your answer to four decimal places.)
c) Would you expect to meet many Asian adult males over 73 inches? Explain why or why not, and justify your answer numerically. YES or NO because the probability that an Asian male is over 73 inches tall is ?
d) The middle 40% of heights fall between what two values?
Write the probability statement.
P(x1 < X < x2) =
State the two values. (Round your answers to one decimal place.)
x1 = x2 =Explanation / Answer
Q1.
a.
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 ) = 66
standard Deviation ( sd )= 2.5
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 64) = (64-66)/2.5
= -2/2.5 = -0.8
= P ( Z <-0.8) From Standard Normal Table
= 0.2119
P(X < 68) = (68-66)/2.5
= 2/2.5 = 0.8
= P ( Z <0.8) From Standard Normal Table
= 0.7881
P(64 < X < 68) = 0.7881-0.2119 = 0.5763
c.
P(X > 73) = (73-66)/2.5
= 7/2.5 = 2.8
= P ( Z >2.8) From Standard Normal Table
= 0.0026
no, it is an outlier
d.
LESSAR THAN
P ( Z < x ) = 0.3
Value of z to the cumulative probability of 0.3 from normal table is -0.5244
P( x-u/s.d < x - 66/25 ) = 0.3
That is, ( x - 66/25 ) = -0.5244
--> x = -0.5244 * 25 + 66 = 52.89
GREATER THAN
P ( Z > x ) = 0.3
Value of z to the cumulative probability of 0.3 from normal table is 0.5244
P( x-u / (s.d) > x - 66/25) = 0.3
That is, ( x - 66/25) = 0.5244
--> x = 0.5244 * 25+66 = 79.11
Middle outlier be = P[ 0.30 < X < 0.70 ] = P( X < 0.30 ) < X < P( X > 0.30 ) = [ 52.89, 79.11]
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