A digital IC manufacturer has determined that for adequate yield, a specification on the drive current for a minimum geometry MOSFET device must be in the range from 100mA/um < k < 200mA/um. Therefore, LSL=100mA/um and USL = 200mA/um. Suppose that to yield a working IC, the transistor yield (i. e.those that pass the spec) must be 99.98%.How manystandard deviations from the mean must LSL and USLrepresent?Assume that the process mean is equal to thedesired mean

The standard deviation from the mean of USL and LSL is 50

Explanation:

Since LSL= 100

USL = 200

STEP1: Find the mean value;

To find the mean, we take the average of LSL and USL.

200 + 100 = 300

Therefore;

300 ÷ 2= 150

STEP2: Find the variance;

To find the variance, we first minus the mean value from the values of ULS and LSL and then square the value

(100 - 150)^2 = 2500

(200 - 150)^2 = 2500

To get the variance

Then sum up the values and divide by the number of values which is 2

(2500+2500) ÷ 2 = 2500

STEP3 : Find the standard deviation;

Standard deviation= √variance

Standard deviation = √2500 = 50

the answer is b ) square root of 5

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft