The width of a tool used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micron and a standard deviation of 0.05 micron. What is the probability that a randomly selected tool will have a width between 0.47 and 0.63 microns? What is a specific width value so that 90% of the tools are below that value?

Step-by-step explanation:

Since the width of a tool used for semiconductor manufacturing is assumed to be normally distributed,

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = width of selected tools.

µ = mean width

σ = standard deviation

From the information given,

µ = 0.5 micron

σ = 0.05 micron

We want to find the probability that a randomly selected tool will have a width between 0.47 and 0.63 microns. It is expressed as

P(0.47 ≤ x ≤ 0.63)

For x = 0.47,

z = (0.47 - 0.5)/0.05 = - 0.6

Looking at the normal distribution table, the probability corresponding to the z score is 0.27

For x = 0.63,

z = (0.63 - 0.5)/0.05 = 2.6

Looking at the normal distribution table, the probability corresponding to the z score is 0.9953

P(0.47 ≤ x ≤ 0.63) = 0.9953 - 0.27 = 0.7253

Second question

90% = 90/100 = 0.9

Looking at the normal distribution table, the z score corresponding to 90% is 1.285

Therefore

1.285 = (x - 0.5)/0.05

1.285 × 0.05 = x - 0.5

0.06425 = x - 0.5

x = 0.5 + 0.06425

x = 0.56425

The width value is approximately 0.56

.5 lb

step-by-step explanation:

12 in / 6 in =2 in

.25 x 2 = .5 lb

give more details

step-by-step explanation: