Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 11 ounces. a. The process standard deviation is .20 ounces, and the process control is set at plus or minus standard deviations. Units with weights less than or greater than ounces will be classified as defects. What is the probability of a defect (to 4 decimals)

The question supplied is incomplete :

The x parameters aren't given ; "Units with weights less than or greater than ounces will be classified as defects"

Assume the unit weights Given are ;

Units with weights less than 10.8 or greater than 11.2 ounces will be classified as defects

Just follow the procedure in the solution for any value of unit weight interval given.

0.3173

Step-by-step explanation:

Mean weight, m = 11 ounces

Standard deviation, s = 0.2 ounces

Find the Zscore for each unit weight :

Z = (x - mean) / standard deviation

P(x < 10.8) :

Z = (10.8 - 11) / 0.2 = - 1

P(Z < - 1) = 0.15866

P(x > 11.2) :

Z = (11.2 - 11) / 0.2 = 1

P(Z > 1) = 0.84134

P(x < 11.2) - P(x < 10.8) = 1 - (P(Z < 1) - P(Z < - 1)) = 1 - 1 - (0.84134 - 0.15866) = 1 - 0.68268 = 0.31732

Hence, Probability of a defect is 0.3173

relating to or denoting a system of numbers and arithmetic based on the number ten, tenth parts, and powers of ten.

it alot will look like this 4.4

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.