Question

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. 

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. 

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. 

Answer 1

\(\mathrm{P}(102<\mathrm{x}<122)=\mathrm{P}[(102-100) / 15)<(\mathrm{x}-\mu) / \sigma<(122-100) / 15)]\)

\(=\mathrm{P}(0.13

$$ \begin{array}{l} =\mathrm{P}(\mathrm{z}<1.47)-\mathrm{P}(\mathrm{z}<0.13) \\ =0.9292-0.5517 \end{array} $$

\(=0.3775\)

The area of the shaded region is 0.3775

Using standard normal table,

\(P(Z>z)=0.1587\)

$$ 1-P(Z

\(P(Z

\(P(Z<1.00)=0.8413\)

\(z=1.00\)

Using z-score formula,

\(\mathrm{x}=\mathrm{z}^{*} \sigma+\mu\)

\(x=1.00 * 15+100=115\)

The indicated IQ score is 115

Using standard normal table,

\(P(Z>z)=0.8\)

$$ 1-P(Z

\(P(Z

\(P(Z<-0.84)=0.2\)

\(z=-0.84\)

Using z-score formula,

\(\mathrm{x}=\mathrm{z}^{*} \sigma+\mu\)

\(x=-0.84 * 15+100=87.4\)

he indicated \(1 Q\) score, \(x,\) is 87.4

Answer 2

solution

given that

(A)P(102< x < 122) = P[(102-100) /15 < (x - ) / < (122-102) /15 )]

= P( 0.13< Z < 1.33)

= P(Z <1.33 ) - P(Z <0.13 )

Using z table   

= 0.9082-0.5517

probability= 0.3565

(B)

Using standard normal table,

P(Z > z) = 0.1587

= 1 - P(Z < z) = 0.1587

= P(Z < z) = 1 - 0.1587

= P(Z < z ) = 0.8413

= P(Z < 0.9998) = 0.8413  

z =0.9998 ( using z table )

Using z-score formula,

x = z * +

x = 0.9998 * 15+100

x = 114.9970

x=115

(C)

Using standard normal table,

P(Z > z) = 0.8

= 1 - P(Z < z) = 0.8

= P(Z < z) = 1 - 0.8

= P(Z < z ) = 0.2

= P(Z < -0.84) = 0.2

z =- 0.84 ( using z table )

Using z-score formula,

x = z * +

x = -0.84 * 15+100

x = 87.4