Question

Please answer the questions below

1. Find the​ z-score that has 48.8% of the​ distribution's area to its left?

2. Find the​ z-scores for which 5% of the​ distribution's area lies between minus−z and z?

3. Find the indicated area under the standard normal curve. Between z=0 and z=1.34 The area between

z=0 and z=1.34 under the standard normal curve is?

Answer 1

solution

P(Z < z) = 48.8%

= P(Z < z) = 0.488

= P(Z < -0.03) = 0.488

z =-0.03 Using standard normal z table,

2.

P(-z < Z < z) = 5%=0.05

P(Z < z) - P(Z < -z) = 0.05

2 P(Z < z) - 1 = 0.05

2 P(Z < z) = 1 + 0.05 = 1.05

P(Z < z) = 1.05 / 2 = 0.525

P(Z <0.06) = 0.525

z  ± 0.06 using z table

lies between -0.06 to +0.06

3.

P( 0< Z <1.34 )

= P(Z < 1.34) - P(Z < 0)

Using z table   

= 0.9099-0.5

area=0.4099

z=0 and z=1.34 under the standard normal curve is 0.4099

answered by: anonymous