Question

1 point) Suppose X is a normally distributed random variable with H9 and ơ 1.3. Find each of the following probabilities: (a) P(12 < X < 15) (b) P(6.1 K X K 16.7)- (c) P(11.1 K X K 16.7)- (d) P(X 2 11.5)- (e) P(X s 16.7)-

Answer 1

Solution :

Given that,

mean = $\mu$ = 9

standard deviation = $\sigma$ =1.3

a ) P (12 $\leq$ x $\leq$ 15)

P (12 - 9 / 1.3)  $\leq$ ( x -  $\mu$/ $\sigma$)$\leq$ ( 15- 9 / 1.3)

P ( 3 / 1.3 $\leq$ z $\leq$ 6 / 1.3 )

P ( 2.31 $\leq$ z $\leq$ 4.62 )

P (z $\leq$ 4.62) - p ( z $\leq$ 2.31 )

Using z table

= 1- 0.9896

= 0.0104

Probability = 0.0104

b ) P (6.1$\leq$  x $\leq$ 16.7 )

P (6.1 - 9 / 1.3 ) $\leq$  ( x -  $\mu$/ $\sigma$)$\leq$  (16.7 - 9 / 1.3)

P ( -2.9/1.3$\leq$ z $\leq$ 7.7/1.3)

P ( - 2.23$\leq$  z $\leq$ 5.92 )

P (z $\leq$  5.92)p ( z $\leq$-2.23 )

Using z table

= 1- 0.0129

= 0.9871

Probability =0.9871

c) P (11.1$\leq$  x $\leq$ 16.7 )

P (11.1-9/ 1.3 ) $\leq$  ( x -  $\mu$/ $\sigma$)$\leq$  (16.7 - 9 / 1.3)

P (2.1/3$\leq$ z $\leq$ 7.7/1.3)

P (1.62$\leq$  z $\leq$ 5.92 )

P (z $\leq$  5.92)p ( z $\leq$1.62 )

Using z table

= 1- 0.9474

= 0.0526

Probability =0.0526

d) P ( x $\geq$  11.5 )

= 1 -P( x $\leq$ 11.5 )

= 1 -P ( x -  $\mu$/ $\sigma$)$\leq$  (11.5 - 9 / 1.3)

= 1 -P( z $\leq$ 2.5/1.3)

= 1 -P(z $\leq$ 1.92 )

Using z table

= 1- 0.9726

= 0.0274

Probability =0.0274

e) P ( x $\leq$ 16.7 )

P ( x -  $\mu$/ $\sigma$)$\leq$  (16.7 - 9 / 1.3)

P (z $\leq$ 7.7/1.3)

P ( z $\leq$ 5.92 )

Using z table

= 1

Probability =1