Question

8.- Normal area calculations. Z = Standard NOrmal

a. Determine P(-1.5 < Z < 1.5) (the area beneath the standard normal curve between Z=-1.5 and Z=1.5).

Probability:  

b. Find the P(0.8 < Z < 1.7).

Probability:  

c. Find the value of z such that P(Z < z)=0.25.

Probability:  

d. Your answer to part (c) above is Q₁ for a standard normal variable, since the relative frequency to the left of the value is 0.25. What are Q₂ (the median, Find the value of z such that P(Z < z)=0.50) and Q₃(Find the value of z such that P(Z < z)=0.75)of a standard normal variable?

Q₂:  

Q₃:  

e. Suppose X has a normal distribution with mean 10.0 and standard deviation 5.0.

What is the P(2.0 < X < 14.0)?

P(2.0 < X < 14.0)=  

f. Suppose X has a normal distribution with mean 10.0 and standard deviation 2.0. What is the P(2.0 < X < 14.0)?

P(2.0 < X < 14.0)=

Answer 1

Solution :

8)

a) P(-1.5 < z < 1.5 )

= P(z <1.5 ) - P(z < -1.5 )

Using standard normal table

= 0.9332 - 0.0668 = 0.8664

Probability =0.8664

b) P(0.8 < Z < 1.7)

= P(z <1.7 ) - P(z < 0.8 )

= 0.9554 - 0.7881 = = 0.1673

Probability =0.1673

c) P(Z < z)= 0.25.

z = -0.674

d)

P(Z < z)= 0.50

z = 0

P(Z < z)=0.75

z = 0.674

e)

mean = = 10

standard deviation = = 5

P( 2< x < 14 ) = P[(2 -10)/ 5) < (x - ) /  < (14 -10) /5 ) ]

= P( -1.6 < z < 0.8 )

= P(z < 0.8 ) - P(z < -1.6 )

= 0.7881 - 0.0548 = 0.7333

Probability =0.7333

f)

mean = = 10

standard deviation = = 2

P( 2< x < 14 ) = P[(2 -10)/ 2) < (x - ) /  < (14 -10) /2 ) ]

= P( -4 < z < 2)

= P(z < 2) - P(z < -4 )

= 0.9772 - 0 = 0.9772

Probability =0.9772