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

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step-by-step explanation:

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x