Question

In: Statistics and Probability

QUESTION 5 The variable Z has a standard normal distribution. The probability P(- 0.5 < Z...

QUESTION 5

The variable Z has a standard normal distribution. The probability P(- 0.5 < Z < 1.0) is:

a.

0.5328

b.

0.3085

c.

0.8413

d.

0.5794

QUESTION 6

If a random variable X is normally distributed with a mean of 30 and a standard deviation of 10, then P(X=20) =

a.

0.4772

b.

-0.4772

c.

-2.00

d.

0.00

QUESTION 7

If P( -z < Z < +z) = 0.8812, then the z-score is:

a.

1.56

b.

1.89

c.

0.80

d.

2.54

QUESTION 8

If the mean of a normal distribution is negative,

a.

the standard deviation must also be negative.

b.

the variance must also be negative.

c.

a mistake has been made in the computations, because the mean of a normal distribution can not be negative.

d.

None of these alternatives is correct.

QUESTION 9

The starting salaries of individuals with an MBA degree are normally distributed with a mean of $90,000 and a standard deviation of $20,000. What is the probability that a randomly selected individual with an MBA degree will have a starting salary of at least $78,500?

  1. 0.2810
  2. 0.8840
  3. 0.7190
  4. 0.8210

QUESTION 10

The starting salaries of individuals with an MBA degree are normally distributed with a mean of $90,000 and a standard deviation of $20,000. What is the lowest salary for those individuals with an MBA degree whose starting salary is in the top 25 percent?

  1. $104,500
  2. $103,600
  3. $83,750
  4. $106,400

Solutions

Expert Solution

5)

P(-0.5 < Z < 1) = P(Z < 1) - P(Z < -0.5) = 0.8413 - 0.3085 = 0.5328

6)

mean = = 30

standard deviation = = 10

x = 20

P(x = 20 ) = 0

In continuous  distribution the exactly probability is always equal to 0 .

7)

P(-z < Z < z) = 0.8812

P(Z < z) - P(Z < z) = 0.8812

2P(Z < z) - 1 = 0.8812

2P(Z < z) = 1 + 0.8812

2P(Z < z) = 1.8812

P(Z < z) = 1.8812 / 2

P(Z < z) = 0.9406

P(Z < 1.56) = 0.9406

z score = 1.56

8)

None of these alternatives is correct.

9)

mean = = 90000

standard deviation = = 20000

P(x 78500) = 1 - P(x   78500)

= 1 - P((x - ) / (78500 - 90000) / 20000)

= 1 -  P(z -0.58)  

= 1 - 0.2810

= 7190

Probability = 0.7190

10)

P(Z > z) = 25%

1 - P(Z < z) = 0.25

P(Z < z) = 1 - 0.25

P(Z < 0.67) = 0.75

z = 0.67

Using z-score formula,

x = z * +

x = 0.67 * 20000 + 90000 = 103600

salary = 103600


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