I need help with these two questions thanks 4 E 032 Suppose the force acting on

Question

I need help with these two questions thanks

4 E 032 Suppose the force acting on a standardizing and then using a standard normal curve table from the Appendix Tables. (Round your answers to four decimal places.) that helps to support a building is a normally d by (a) P(X S 16) (b) PX s 18.5) (c) P(X 2 11) (d) P(15 s X s 19) (e) P(ix - 161 s 2) An article suggests that substrate concentration mao 3 of influent to a reactor is normally distributed with -0 60 and 0-0 06 (a) What is the probability that the concentration exceeds 0.70? Round your answers to four decimal places. (b) What is the probability that the n is at most 0.40? mg/cm3 The largest 5% of all concentration values are above You may need to use the appropriate table in the Appendix of Tables to answer this question

Explanation / Answer

Result:

a).

z value for 16 , z=(16-16)/1.25 =0.0

P( x 16) = P( z <0)

=0.5

b).

z value for 18.5 , z=(18.5-16)/1.25 =2.0

P( x 18.5) = P( z <2)

=0.9772

c).

z value for 11 , z=(11-16)/1.25 = -4.0

P( x 11) = P( z >-4.0)

=1.0000

d).

z value for 15 , z =(15-16)/1.25 = -0.8

z value for 19 , z =(19-16)/1.25 = 2.4

P( 15x19) =P( -0.8<z<2.4)

=P( z <2.4) –P( z< -0.8)

=0.9918 - 0.2119

=0.7799

e).

P(|x-16| 2) =P( -2 <z<2)

P( z <2) –P( z< -2)

=0.9772-0.0227

=0.9545

Second part

a).

z value for 0.70 , z =(0.70-0.60)/0.06 = 1.67

P( x>0.70) = P( z >1.67)

= 0.0475

b).

z value for 0.40 , z =(0.40-0.60)/0.06 = -3.33

P( x0.40) = P( z < -3.33)

= 0.0004

c).

z value to top 5% =1.645

x =0.60+1.645*0.06

= 0.6987

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