A manufactured product has a length that is normally distributed with a mean of 12 cm. The product will be unusable if the length is 11 V2 cm or less. Required:
a. If the probability of this has to be less than 0.01, what is the maximum allowable standard deviation?
b. Assuming this standard deviation, what is the probability that the product’s length will be between 11.75 and 12.35 cm?

a) σ  = 0,1612

b) P [ 11,75 < X < 12,35] = 0,92,44   or    92,44 %

Step-by-step explanation:

NOTE: I assume the unusable length s 11,5 and less

a) For a normal distribution, the probability of P = 0,01 corresponds to z(score) =  -3,1 then:

- 3,1 = ( X - μ₀ ) / σ

- 3,1 = (11,5 - 12 )/ σ

-3,1 = - 0,5 / σ

σ  = 0,1612

b) If σ = 0,1612

P [ 11,75 < X < 12,35]

z₁ (score)  = ( 11,75 - 12 ) / 0,1612

z₁  = - 0,25/ 0,1612

z₁  = -1,55

From z table  P [ 11,75 < X] = 0,0606

z₂ (score) = ( 12,35 - 12) / 0,1612

z₂ = 0,35 / 01612

z₂ = 2,17

From z table  P [ X < 12,35] = 0,9850

Finally P [ 11,75 < X < 12,35] = 0,9850 -  0,0606

P [ 11,75 < X < 12,35] = 0,92,44   or    92,44 %

17 1/3

step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b