The thickness of glass sheets produced by a certain process are normally distributed with a mean of μ = 3.25 millimeters and a standard deviation of σ = 0.10 millimeters. What is the value of c for which there is a 99% probability that a glass sheet has a thickness within the interval ? Round your answer to four decimal places.

C = 0.2576

Step-by-step explanation:

The probability values of general normal distribution N(3.25,0.10) can be related to Φ(z) through the relationship ;

P(a ≤ X ≤ b) = Φ((b-3.25)/0.1) - Φ((a-3.25)/0.1)

Specially, for a = ∞, we have;

F(b) = P( X ≤ b) = Φ((b-3.25)/0.1)

Now from the question we are looking for the probability that the glass sheet has a thickness within the interval;{3.25 - c, 3.25 + c} =0.99

Now, applying the equation from P(a ≤ X ≤ b) earlier, we arrive at;

P(3.25 - c ≤ X ≤ 3.25 + c) = Φ((3.25 + c -3.25)/0.1) - Φ((3.25 - c -3.25)/0.1)

= Φ(c/0.1) - Φ(-c/0.1)

= Φ(c/0.1) + Φ(c/0.1) = 2Φ(c/0.1)

Now, we know that P(3.25 - c ≤ X ≤ 3.25 + c) - 1 = 0.99

And so;

2Φ(c/0.1) - 1 = 0.99

Φ(c/0.1) = (1 + 0.99)/2

Φ(c/0.1) = 0.995

From the table attached, we can see that the value of Z at 0.995 is equal to 2.807034

Thus ;

C/0.1 = 2.5758

So, C = 2.5758 x 0.1 = 0.25758

Approximation to 4 decimal places gives; C = 0.2576

a

step-by-step explanation: