Question

The production of a certain polymer fiber follows a normal distribution with a true mean diameter of 30 μm and a standard deviation of 20 μm. (a) Compute the probability of a measured value greater than 80 μm. (b) Compute the probability of a measured value between 60 and 75 μm.

Answer 1

Answer:

Step-by-step explanation:

Since the the production of the polymer fiber follows a normal distribution, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = measured diameter values.

µ = mean diameter

σ = standard deviation

From the information given,

µ = 30 μm

σ = 20 μm

a) The probability of a measured value greater than 80 μm is expressed as

P(x > 80) = 1 - P(x ≤ 80)

For x = 80,

z = (80 - 30)/20 = 2.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.994

P(x > 80) = 1 - 0.994 = 0.006

b) the probability of a measured value between 60 and 75 μm is expressed as

P(60 ≤ x ≤ 75)

For x = 60,

z = (60 - 30)/20 = 1.5

The probability corresponding to the z score is 0.933

For x = 75,

z = (75 - 30)/20 = 2.25

The probability corresponding to the z score is 0.988

Therefore,

P(60 ≤ x ≤ 75) = 0.988 - 0.933 = 0.055