Question

1. For a standard normal distribution, find:

P(-1.36 < z < 1.08)

2. For a standard normal distribution, find:

P(z > 1.42)

3. GPAs at CCCOnline are normally distributed with a mean of 2.44 and a standard deviation of 0.55. Find the zz-score for a GPA of 2.82.

• 0.3091
• 0.6909
• 1.200
• 1.745
• 0.4909
• 0.7091

Answer 1

solution

P(-1.36 < z < 1.08)

= P(Z <1.08 ) - P(Z <-1.36 )

Using z table,  

= 0.8599-0.0869

=0.7730

probbility=0.7730

(B)  P(z > 1.42)=1 - P(z <1.42 )

Using z table,
= 1 -0.9222

= 0.0778

area = 0.0778

(C)Solution :

Given ,

mean = = 2.44

standard deviation = = 0.55

x=2.82

using z-score formula

z =X -/

z=2.82-2.44/0.55

z=0.6909