Question

Let z denote a variable that has a standard normal distribution. Determine the value z* to satisfy the conditions below. (Round all answers to two decimal places.) (a) P(z < z*) = 0.0256 z* = (b) P(z < z*) = 0.0098 z* = (c) P(z < z*) = 0.0493 z* = (d) P(z > z*) = 0.02 z* = (e) P(z > z*) = 0.0098 z* = (f) P(z > z* or z < -z*) = 0.1974 z* =

Answer 1

Answer:

a) $z* =-1.95$

b) $z* =-2.33$

c) $z* =-1.65$

d) $z* =2.05$

e) $z* =2.33$

f) $z* =1.29$

$-z* =-1.29$

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

a)  $P(Z$

For this case we need to find a value on the standard normal distribution that accumulates 0.0256 of the area on the left tail and we can use the following excel code: "=NORM.INV(0.0256,0,1)" and we got:

$z* =-1.95$

b) $P(Z$

For this case we need to find a value on the standard normal distribution that accumulates 0.0098 of the area on the left tail and we can use the following excel code: "=NORM.INV(0.0098,0,1)" and we got:

$z* =-2.33$

c) $P(Z$

For this case we need to find a value on the standard normal distribution that accumulates 0.0493 of the area on the left tail and we can use the following excel code: "=NORM.INV(0.0493,0,1)" and we got:

$z* =-1.65$

d) $P(Z>z*) =0.02$

For this case we need to find a value on the standard normal distribution that accumulates 0.02 of the area on the right tail and we can use the following excel code: "=NORM.INV(1-0.02,0,1)" and we got:

$z* =2.05$

e)  $P(Z>z*) =0.0098$

For this case we need to find a value on the standard normal distribution that accumulates 0.0098 of the area on the right tail and we can use the following excel code: "=NORM.INV(1-0.0098,0,1)" and we got:

$z* =2.33$

f) $P(Z> z* U Z$

For this case we need to find a value on the standard normal distribution that accumulates 0.1974/2=0.0987 of the area on the right tail and we can use the following excel code: "=NORM.INV(1-0.0987,0,1)" and we got:

$z* =1.29$

$-z* =-1.29$