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 %

898.1666667

step-by-step explanation:

5389/6 = 898.1666667

150.6

all i did was i multiplied 5.02 by 30