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version 3 Questions 06 - 08 Intelligence Quotient (IQ) is a standardized test score designed to assess intelligence. The IQ score is standardized so that the mean is 100 and the standard deviation is 15. Assume that the standardized IQ score follows a normal distribution. Q06 What is the probability that the IQ is between 82 and 109? (a) 70.738% (b) 61.068% (c) 53.325% (d) 46.618% (e) 37.971% Answer (b) Q07 Suppose that a company does not hire anyone whose IQ level is lower than 82. What is the probability that a randomly chosen person is denied by the company due to a low IQ? (a) 11.507% (b) 15.866% (c) 2.275% (d) 28.670% (e) 8.703% Answer (a) Q08 Suppose that there is a society open to people whose IQ score is in the top 9%. What is the required IQ score to join this organization? (a) 117.81 126.25 (e) 124.11 (d) 120.10 (e) 130.87 Answer (d)

Answer 1

• Z-Score is the number of Population standard deviations (σ) from the population mean (μ) a IQ (X) is. It is a measure of how many Population standard deviations (σ) below or above the population mean (μ) a IQ (X) is.
• Mathematically,

            Z Score =

х-и о

           Here we given for : mean (μ) = 100 ; standard deviations (σ) = 15

(6) probability that the IQ is between 82 and 109

• So, Z Score for the IQ (82<X<109) will be as follow:

Here X = 82

               Z score (X= 82) =  (X-μ)/σ

                                              =  (82-100)/15

= -1.2

• From z left table prob(z<-1.2)= 0.1151       

And for  X = 109

               Z score (X= 109) =  (X-μ)/σ

                                      =  (109-100)/15

= 0.6

• From z left table prob(z<0.6)= 0.7257

So, Z Score for the IQ (82<X<109) = [(z left table prob(z<0.6)) - (z left table prob(z<-1.2))]

•    Z Score for the IQ(82<X<109) = 0.7257 - 0.1151       

                                                               = 0.6106

• Z Score for the IQ(82<X<109) = 0.6106

                                                             = 0.6106*100

                                                            = 61.06%

(7) prob. That company does not hire anyone whose IQ level is lower than 82

• So, Z Score for the IQ (82<X) will be as follow:

Here  X = 82

               Z score (X= 82) =  (X-μ)/σ

                                         =  (82-100)/15

= -1.2

From z left table prob(z<-1.2)= 0.115099

en

• Z Score for the IQ (82<X)= 0.115099*100

= 11.509 % .................Answer

(8) given people whose IQ score is in the top 9% = 9/100 = 0.09

From z right table Prob(z>0.09) = 1.341          . . . . . . . . . . . . . . . . . . (i)

We know

Z-score =  (X-μ)/σ                (from equation (i) z=1.341)

1.341 =  (X-100)/15

(1.341)*(15) = (X-100)

(X-100) = (1.341)*(15)

X = 20.115 + 100

X = 120.115