Question

6.2.10-T Question Help Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 The indicated IQ score is (Round to the nearest whole number as needed) Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of the table. Click to view page 2 of the table. The indicated I score x S (Round to one decimal place as needed) Find the area of the shaded region. The graph to the night depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. 102130 The area of the shaded region is (Round to four decimal places as needed) Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of the table. Click to view page 2 of the table The area of the shaded region is (Round to four decimal places as needed)

Answer 1

X N (mean = , variance = o)

So,

Z-A-MN(0,1)

Thus,

   ( we have to use standard normal table or software to find this value, we will use R Studio here, code: pnorm(z)

(2) = P(Z < 2) = cdf of Z

So,

( for this also we have to use standard normal table or software to find this value, we will use R Studio here, code: qnorm(z) )

(2) = k→ = 0-1(k) = kth quantile of Z

I will mention the codes along with output below

X: IQ scores

a)

P(X>) = 0.4470 1- P(X<x) = 0.4470 = P(x < 2) = 0.553 =P473100 < 45300) = 0.553= *4*300) = 0,553 = 47100 = 0–(0.553) = 0.1332445

R code: qnorm(0.553)

So, x= 101.9987 = 102

b)

P(X<I) = 0.75 PYT100<---7300) = 0.75 = ""7300) = 0.75 100 =0-1(0.75) = 0.6744898

R code: qnorm(0.75)

So, x = 110.1173 == 110.1

c)

P(102 < X < 130) = P(X < 130) - P(X < 102) = P4-100 < 130 - 100) – P4X – 100 < 102 – 100 = P(Z < 130 = 100) – P(Z < 102 – 100, = N(2) – che = 0.9772499 -0.5530351

R Code: pnorm(2) , pnorm(2/15)

= 0.4242

d)

P(70 < X < 105) = P(X < 105) – P(X < 70) = P(X – 100 105 – 100, X - 100 =P 15 *** <100 - 109) - PA100, 0 - 100. 15 15 15 = P(Z < 105 = 100) – P(Z < 70 – 100, = 0) – P(-2) = 0,6305587 – 0.02275013 15 ) 15

R Code: pnorm(1/3) , pnorm(-2)

= 0.6078

R Studio Outputs:

> qnorm(0.553)
[1] 0.1332445
> qnorm(0.75)
[1] 0.6744898
> pnorm(2)
[1] 0.9772499
> pnorm(2/15)
[1] 0.5530351
> pnorm(1/3)
[1] 0.6305587
> pnorm(-2)
[1] 0.02275013