Question

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case find the probability of the given scores.

*Round your answers to four decimal places.

1. Find the probability of a score less than -2.04

2. Find the probability of a score less than 2.33

3. Find the probability of a score greater than 0.82

4. Find the probability of a score greater than -1.50

5. Find the probability of a score between -2.75 and -2.00

6. Find the probability of a score between -2.20 and 2.50

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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

1.
the probability of a score less than -2.04
P(X < -2.04) = (-2.04-0)/1
= -2.04/1= -2.04
= P ( Z <-2.04) From Standard Normal Table
= 0.0207

2.
the probability of a score less than 2.33
P(X < 2.33) = (2.33-0)/1
= 2.33/1= 2.33
= P ( Z <2.33) From Standard Normal Table
= 0.9901

3.
the probability of a score greater than 0.82
P(X > 0.82) = (0.82-0)/1
= 0.82/1 = 0.82
= P ( Z >0.82) From Standard Normal Table
= 0.2061

4.
the probability of a score greater than -1.50
P(X > -1.5) = (-1.5-0)/1
= -1.5/1 = -1.5
= P ( Z >-1.5) From Standard Normal Table
= 0.9332

5.
the probability of a score between -2.75 and -2.00
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -2.75) = (-2.75-0)/1
= -2.75/1 = -2.75
= P ( Z <-2.75) From Standard Normal Table
= 0.003
P(X < -2) = (-2-0)/1
= -2/1 = -2
= P ( Z <-2) From Standard Normal Table
= 0.0228
P(-2.75 < X < -2) = 0.0228-0.003 = 0.0198

6.
the probability of a score between -2.20 and 2.50
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -2.2) = (-2.2-0)/1
= -2.2/1 = -2.2
= P ( Z <-2.2) From Standard Normal Table
= 0.0139
P(X < 2.5) = (2.5-0)/1
= 2.5/1 = 2.5
= P ( Z <2.5) From Standard Normal Table
= 0.9938
P(-2.2 < X < 2.5) = 0.9938-0.0139 = 0.9799