Question

Normal Distribution Problems Include any pertinent information used in your calculations: z-scores, table areas, etc. In other words, briefly describe how you came up with your answer.

4. A bag of potato chips claims to contain 7 oz. of potato chips. Random sampling determines that the bags contain an average of 7.2 oz. with a standard deviation of 0.081 oz.

a. What is the maximum weight that 20% of the bags contain less than?

b. What is the minimum weight that the heaviest 15% of the bags contain?

c. In what weight interval are the middle 60% of the bags contained?

Answer 1

A) P(X < x) = 0.2

Or, P((X - )/ < (x - )/) = 0.2

Or, P(Z < (x - 7)/0.081) = 0.2

Or, (x - 7)/0.081 = -0.84

Or, x = -0.84 * 0.081 + 7

Or, x = 6.932

B) P(X > x) = 0.15

Or, P((X - )/ > (x - )/) = 0.15

Or, P(Z > (x - 7)/0.081) = 0.15

Or, P(Z < (x - 7)/0.081) = 0.85

Or, (x - 7)/0.081 = 1.04

Or, x = 1.04 * 0.081 + 7

Or, x = 7.08

C) P(X < x) = 0.2

Or, P((X - )/ < (x - )/) = 0.2

Or, P(Z < (x - 7)/0.081) = 0.2

Or, (x - 7)/0.081 = -0.84

Or, x = -0.84 * 0.081 + 7

Or, x = 6.932

P(X > x) = 0.2

Or, P((X - )/ > (x - )/) = 0.2

Or, P(Z > (x - 7)/0.081) = 0.2

Or, P(Z < (x - 7)/0.081) = 0.8

Or, (x - 7)/0.081 = 0.84

Or, x = 0.84 * 0.081 + 7

Or, x = 7.068

So the middle 60% lies in the interval 6.932 and 7.068