Question

Stanford–Binet IQ Test scores are normally distributed with a mean score of 100 and a standard deviation of 11. (b) Write the equation that gives the z score corresponding to a Stanford–Binet IQ test score. z = (x – 100 ) / 11 (c) Find the probability that a randomly selected person has an IQ test score. (Round your answers to 4 decimal places.) 1. P(x > 134) .001 2. P(x < 80) .0345 3. P(84 < x < 116) .9271 − .0729 = .8542 4. P(-1.82 < z < 1.82) 1 (d) Suppose you take the Stanford–Binet IQ Test and receive a score of 156. What percentage of people would receive a score higher than yours? (Round your answer to 2 decimal places.) P .01 %

Answer 1

Let X be the Stanford- Binet IQ Test scores

X~ Normal (100, 11)

(a) Z-score =   

(b) 1. P( X> 134) = P( > )

= P( z > 3.09)

= 1- P( z < 3.09)

= 1- 0.99900

= 0.001

2. P( X < 80) = P(   <   )

= P( z < -1.82)\

= 1- P( z < 1.82)

= 1- 0.96562

= 0.0344

3. P( 84 < X < 116) = P( < <   )

= P( -1.45 < z < 1.45 )

= P( z < 1,45)- P( z < -1.45)

= P( z < 1.45) - 1 + P( z < 1.45)
= 2 P( z < 1.45) - 1

= 2 *0.92647 - 1

= 0.8529

4. P( -1.82 < z < 1.82) = P( z < 1.82) - P( z < -1.82)

= P( z< 1.82) - 1 + P( z < 1.82)

= 2 P( z < 1.82) - 1

= 2 * 0.96562 - 1

= 0.93124

(c)   P( X > 156) = P( > )

= P( z > 5.09)

= 1- P( z < 5.09)

= 1- 0.9999

= 0.0001

= 0.01%