Question

The Intelligence Quotient (IQ) test scores are normally distributed with a mean of 100 and a standard deviation of 15.

What is the probability that a person would score 130 or more on the test?

A..0200

B..0500

C..0228

D..0250

What is the probability that a person would score between 85 and 115?

A..6826

B..6800

C..3413

D..6587

Suppose that you enrolled in a class of 36 students, what is the probability that the class’ average IQ exceeds 130?

A.

almost zero

B.

.0250

C.

.0500

D.

.2280

What is the probability that a person would score between 115 and 130?

	A.	.1587
	B.	.1359
	C.	.0228
	D.	.3200

Answer 1

Solution:

Given that,

mean = = 100

standard deviation = =15

1 ) p ( x > 130 )

= 1 - p (x < 130 )

= 1 - p ( x -  / ) < ( 130 - 100 / 15)

= 1 - p ( z < 30 / 15 )

= 1 - p ( z < 2)

Using z table

= 1 - 0.9772

= 0.0228

Probability = 0.0228

Option C is correct.

2 ) p (85 < x < 115 )

= p (85-100/15) < ( x -  / ) < ( 115 - 100 / 15)

= p (-15 / 15 < z < 15 / 15 )

= p (-1 < z < 1)

= p (z < 1 ) - p ( z < -1)

Using z table

= 0.8413 - 0.1587

= 0.6826

Probability = 0.6826

Option A is correct.

3 ) n = 36

So,

   = 100

=  ( /n) = (15 / 36 ) = 2.5

( 130 )

= 1 - p (    130 )

= 1 - p ( - /)   (130 - 100 / 2.5 )

= 1 - p( z   30 / 2.5 )

= 1 - p ( z 12 )   

Using z table

= 1 - 1

= 0

Probability = 0

Option A is correct.

4) p (115 < x < 130 )

= p (115-100/15) < ( x -  / ) < ( 130 - 100 / 15)

= p (15 / 15 < z < 30 / 15 )

= p (1 < z < 2)

= p (z < 2 ) - p ( z < 1)

Using z table

= 0.9772 - 0.8413

= 0.1359

Probability = 0.1359

Option B is correct.