a certain componnt for newly developed electronic disel engine is considered to be defective if its diameter is less than 8.0 mm or greater than 10.5mm. the distribution of the diameters of these parts is known to be normal with mean of 9.0 mm and standard deviation of 1.5 mm. if a component is randomly selected what is probability that it will be defective?

The probability is 0.58992 or simply 0.59

Step-by-step explanation:

We start by getting the z-scores of the diameter

we have this as follows;

Mathematically;

z-score = (x-mean)/SD

mean = 9.0 mm

SD = 1.5 mm

For 8.00 mm, we have

z = (8-9)/1.5 = -0.67

For 10.5 mm, we have

z = (10.5-9)/1.5 = 1

So the probability we want to calculate is;

P( 1 < x < -0.67)

we can calculate this using the standard normal distribution table

We have this as ;

P( x < -0.67) - P( x > 1) = 0.58992

i think it's either c or d

step-by-step explanation: the entire question is extremely odd. 04 obviously refers to the interest rate. honestly i will say the answer is c. because the (2 x 12) part would mean it's compounded semiannually. i'll say choose c.