Question

An article suggests that substrate concentration (mg/cm³) of influent to a reactor is normally distributed with

μ = 0.60

and

σ = 0.07.

(Round your answers to four decimal places.)

(a)

What is the probability that the concentration exceeds 0.80?

(b)

What is the probability that the concentration is at most 0.50?

(c)

How would you characterize the largest 5% of all concentration values?

The largest 5% of all concentration values are above  mg/cm³.

Answer 1

Solution :

Given ,

mean = = 0.60

standard deviation = = 0.07

P(x >0.80 ) = 1 - P(x<0.80 )

= 1 - P[(x -) / < (080-0.60) / 0.07]

= 1 - P(z <2.86 )

Using z table

= 1 - 0.9979

= 0.0021

probability= 0.0021

(B)

P(X< 0.50) = P[(X- ) / < (0.50-0.60) /0.07 ]

= P(z <-1.43 )

Using z table

= 0.0764

probability=0.0764

(C)

Using standard normal table,

P(Z > z) = 5%

= 1 - P(Z < z) = 0.05

= P(Z < z ) = 1 - 0.05

= P(Z < z ) = 0.95

= P(Z < 1.64 ) = 0.95  

z =1.64 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= 1.64*0.07+0.60

x= 0.7148