Given a standardized normal distribution (with a mean of 0 and a standard deviat

Question

Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1, complete parts (a) through (d). | Click here to view page 1 of the cumulative standardized normal distribution table. EEB Click here to view page 2 of the cumulative standardized normal distribution table a. What is the probability that Z is less than 1.56? The probability that Z is less than 1.56 is Round to four decimal places as needed.) b. What is the probability that Z is greater than 1.83? The probability that Z is greater than 1.83 is (Round to four decimal places as needed.) c. What is the probability that Z is between 1.56 and 1.83? The probability that Z is between 1.56 and 1 83 is (Round to four decimal places as needed.) d. What is the probability that Z is less than 1.56 or greater than 1.83? The probability that Z is less than 1.56 or greater than 1.83 is Round to four decimal places as needed.)

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
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 0
standard Deviation ( sd )= 1
a.
P(X < 1.56) = (1.56-0)/1
= 1.56/1= 1.56
= P ( Z <1.56) From Standard Normal Table
= 0.9406

b.
P(X > 1.83) = (1.83-0)/1
= 1.83/1 = 1.83
= P ( Z >1.83) From Standard Normal Table
= 0.0336

c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1.56) = (1.56-0)/1
= 1.56/1 = 1.56
= P ( Z <1.56) From Standard Normal Table
= 0.9406
P(X < 1.83) = (1.83-0)/1
= 1.83/1 = 1.83
= P ( Z <1.83) From Standard Normal Table
= 0.9664
P(1.56 < X < 1.83) = 0.9664-0.9406 = 0.0258

d.
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 1.56) = (1.56-0)/1
= 1.56/1= 1.56
= P ( Z <1.56) From Standard Normal Table
= 0.9406
P(X > 1.83) = (1.83-0)/1
= 1.83/1 = 1.83
= P ( Z >1.83) From Standard Normal Table
= 0.0336
P( X < 1.56 OR X > 1.83) = 0.9406+0.0336 = 0.9742

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