Question

***PLEASE SHOW HOW TO SOLVE IN EXCEL*** NOT HANDWRITTEN

Given that z is a standard normal random variable, find z for each situation.

a. The area to the left of z is .9750.

b. The area between 0 and z is .4750.

c. The area to the left of z is .7291.

d. The area to the right of z is .1314.

e. The area to the left of z is .6700.

f. The area to the right of z is .3300.

Answer 1

z is a standard normal variable.

a.The area to the left of z is 0.9750

Therefore,

Z=1.96 ( since using Excels (=NORMSINV(0.9750)))

# calculated value of z for the area value to the left of 0.9750 is 1.96

b. The area between 0 and z is .4750.

ANS

From the information, observe that z is a standard normal variable.

The area between 0 and z is 0.4750

Therefore,

P(0<Z<z)=0.4750

P(Z<z)-P(Z<0)=0.4750

P(Z<z)=P(Z<0)+0.4750

P(Z<z) =0.5+0.4750    ieP(Z<0)=0.5

  P(Z<z)=9750

Therefore,

Z=1.96 ( since using Excels (=NORMSINV(0.9750)))

The calculated value of z for the area between 0 and z is 0.4750 is 1.96

c. The area to the left of z is .7291.

Ans:

From the information, observe that z is a standard normal variable.

The area to the left of z is 0.7291

Therefore,

Z=0.61( since using Excels (=NORMSINV(0.7291)))

# calculated value of z for the area value to the left of 0.7291 is 0.61

d. The area to the right of z is .1314.

Ans:

From the information, observe that z is a standard normal variable.

The area to the right of z is 0.1314

Therefore,

P(Z>z)=0.1314

1-P(Z<z)=0.1314

P(Z<z)=1-0.1314=0.8686

Z=1.12( since using Excels (=NORMSINV(0.8686)))

Therefore,

Part d

The calculated value of z for the area value to the right of 0.1314 is 1.12

e. The area to the left of z is .6700.

Ans:

From the information, observe that z is a standard normal variable.

The area to the left of z is 0.6700

Therefore,

P(Z<z)=0.6700

Z=0.44( since using Excels (=NORMSINV(0.6700)))

Therefore,

Part e

#The calculated value of z for the area value to the left of 0.6700 is 0.44

f. The area to the right of z is .3300.

Ans:

P(Z>z)=0.3300

1-P(Z<z)=0.3300

P(Z<z)=1-0.3300=0.6700

Therefore,

Z=0.44( since using Excels (=NORMSINV(0.6700)))

#The calculated value of z for the area value to the right of 0.3300 is 0.44