The amount of coffe in a can has a mean of 350g and a standard devation of 4g

Question

What percent of cans have less than 338g of coffee?

What is the probability that a can has between 342g and 350g of coffee?

What is the probability that can has less than 342g or more than 346g of coffee?

Answer 1

1) What percent of cans have less than 338g of coffee?

Z-score = (x - μ) / σ = (338 - 350) / (4) = -3

For a z-score of -3, P(x<Z) = 0.0013499

This means that less than 0.13% of cans have less than 338 g of coffee.

2) What is the probability that a can has between 342g and 350g of coffee?

Here we calculate two z-scores and their respective probabilities, and then we subtract the difference between these two probabilities to find the likelihood.

1st Z-score = (x - μ) / σ = (342 - 350) / (4) = -2 --> P(x < Z) = 0.02275

2nd Z-score = (x - μ) / σ = (350 - 350) / (4) = 0 --> P(x < Z) = 0.5

Probability that a can has between 342g and 350g = (0.5) - (0.02275) = 0.47725 or approximately 0.48

3) What is the probability that can has less than 342g or more than 346g of coffee?

P(x < 342) + P(x > 346)

1st Z-score: (x - μ) / σ = (342 - 350) / (4) = -2 --> P(x < Z) = 0.02275

2nd Z-score: (x - μ) / σ = (346 - 350) / (4) = -1 --> P(x > Z) = 0.84134

(0.02275) + (0.84134) = 0.86409 or approximately 0.86

Answer 2

mean = 350 SD = 4

x = weight of coffee

P(x < 338)

z = (x - mean)/SD

Compute z for 338: z = (338 = 350)/4 = - 3

P(x < 338) = P(z < = - 3) = 0.0013

P(342 < x < 350)

z = (342-350)/4 = -2

z = (350 - 350)/4 = 0

P(342 < x < 350) = P(-2 < z < 0) = P(z < 0) - P(z < -2) = 0.5 - 0.228 = 0.272

P(x < 342) or P(x > 346) = 1 - P(342 < x < 346)

for 342 z = -2

for 346, z = (346-350)/4 = -1

1 - P(342 < x < 346) = 1 - P(-2 < z < -1) = 1 - (P(z<-1) - P(z < -2)) = 1 - (0.1587 - 0.0228) = 1 - 0.1359 = 0.8641