Question

Employees of a local university have been classified according to gender and job type.

	Male	Female
Faculty	110	60
Salaried Staff	50	60
Hourly Staff	80	40

If an employee is selected at random, what is the probability that the employee is a member of the hourly staff given that the employee is male?

		0.124
		0.235
		0.260
		0.286
		0.333

If the scores on an aptitude test are normally distributed with mean 600 and standard deviation 50, what proportion of the test scores are less than 675?

		0.1977
		0.8500
		0.9332
		0.8023

Consider a normal population with a mean of 10 and a variance of 2. Find P(X > 6).

		0.0668
		0.9332
		0.8413
		0.9772

Consider a normal population with a mean of 10 and a standard deviation of 2. Find P(X > 11).

		0.0228
		0.3085
		0.8413
		0.9772

Answer 1

Solution :

from given table,

The probability that the employee is a member of the hourly staff given that the employee is male ,

= 80 / (110 + 50 + 80)

= 0.333

P(x < 675) = P[(x - ) / < (675 - 600) / 50]

= P(z < 1.5)

= 0.9332

6)

P(x > 6) = 1 - P(x < 6)

= 1 - P[(x - ) / < (6 - 10) / 2]

= 1 - P(z < -2)

= 0.9772

7)

P(x > 11) = 1 - P(x < 11)

= 1 - P[(x - ) / < (11 - 10) / 2]

= 1 - P(z < 0.5)

= 0.3085