Question

1. Scores on an endurance test for cardiac patients are normally distributed with a mean of 182 and a standard deviation of 24. What is the probability that a patient has a score above 170?

0.1915

0.3085

0.4505

0.6915

2.

The number of interviews received within a month by TSU business graduates is normally distributed with a mean of 8 interviews and a standard deviation of 2 interviews. 75% of all graduates receive more that how many interviews?

7.45

8.31

6.65

5.87

3. The number of interviews received within a month by TSU business graduates is normally distributed with a mean of 8 interviews and a standard deviation of 2 interviews. 15% of all graduates receive no more than how many interviews?

6.47

5.93

7.84

8.31

4. The number of interviews received within a month by TSU business graduates is normally distributed with a mean of 8 interviews and a standard deviation of 2 interviews. 45% of all graduates receive more than how many interviews?

10.11
		11.92
		8.25
		9.43

Answer 1

Solution :

Given that ,

1) mean = = 182

standard deviation = = 24

P(x > 170) = 1 - p( x< 170)

=1- p P[(x - ) / < (170 - 182) / 24]

=1- P(z < -0.5 )

Using z table,

= 1 - 0.3085

= 0.6915

2) Given that,

mean = = 8

standard deviation = = 2

Using standard normal table,

P(Z > z) = 75%

= 1 - P(Z < z) = 0.75

= P(Z < z) = 1 - 0.75

= P(Z < z ) = 0.25

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 2 + 8

x = 6.65

3) Using standard normal table,

P(Z < z) = 15%

= P(Z < z) = 0.15

= P(Z < -1.036) = 0.15   

z = -1.036

Using z-score formula,

x = z * +

x = -1.036 * 2 + 8

x = 5.93

4) Using standard normal table,

P(Z > z) = 45%

= 1 - P(Z < z) = 0.45

= P(Z < z) = 1 - 0.45

= P(Z < z ) = 0.55

= P(Z < 0.1257 ) = 0.55

z = 0.1257

Using z-score formula,

x = z * +

x = 0.1257 * 2 + 8

x = 8.25