Question

hey im having trouble with this

Use Excel to find the Standard Normal Probability for the following questions. Don't forget to sketch the normal distribution and shade the required area.

a) What is the area under the standard normal curve to the left of z = 0.89? (4dp) Answer

b) What is the area under the standard normal curve between z = -0.89 and z = 1.63? (4dp) Answer

c) What is the z-value that gives the right hand tail area equal to 0.0268? (2dp) Answer d) What is the absolute value of z such that the total area under the standard normal curve between –z and +z will be 0.9443? (2dp) Answer

Answer 1

In an Excel spreadsheet, the function

=NORMSDIST(z)

gives area under the standard normal curve ( probability for the interval from –∞ to z ).

a)

Given z = 0.89

Using excel formula,

=NORMSDIST(0.89)

0.8133

( round upto four decimals place )

The area under the standard normal curve to the left of z = 0.89 is 0.8133

b)

Given

z1 = -0.89

z2 = 1.63

Using excel formula,

=NORMSDIST(1.63)-NORMSDIST(-0.89)

0.7617

( round upto four decimals place )

The area under the standard normal curve between z = -0.89 and z = 1.63 is 0.7617

c)

Given

right hand tail area = 0.0268

Area to the left of z-score = 1 - 0.0268 = 0.9732

Using excel formula,

=NORMSINV(0.9732)

1.93

( round upto two decimals place )

The z-value that gives the right hand tail area equal to 0.0268 is 1.93

d)

Given  the total area under the standard normal curve between –z and +z = 0.9443

The area which is outside the given range = 1 - 0.9443 = 0.0557.

Since the graphs are symmetrical, the area to the left of -z score = 0.0557/2 = 0.0279.

Also area to the right of z-score = 0.0279

Now

Using excel formula,

=NORMSINV(0.0279)

-1.91

( round upto two decimals place )

i.e

The absolute value of z such that the total area under the standard normal curve between –z and +z will be 0.9443

is 1.91