A manufacturing process produces potato chip bags that have Normally distributed weights, with a mean weight of 15 oz. and a standard deviation of .3 oz. What is the probability that 16 randomly selected bags have a mean weight that exceeds 15.2 oz?

This a normal distribution. We need to calculate the Z score, and then with a Z-table, find the probability. We do that, using the mean, standard deviation and the value from the question 15.2 oz.

So, Z score: [tex]Z=\frac{X-u}{o} =\frac{15.2-15}{0.3}=0.6[/tex]

Refer to the image attached. If we consult the z-table, we find that 0.6 belongs to a probability of 0.73. But, we need to find the probability of weight that exceeds 15.2 oz. So, we need to subtract the 0,73, from the total probability, which is one:

[tex]P=1-0.73 =0.27[/tex]

Just, remember that 0.73 belongs to a probability where the weight is less than 15.2, that's why we need to subtract.

Hence, the probability is 27%.

raphealnwobi raphealnwobi · Answer 2 · 2022-03-08T12:15:25+01:00

The probability that 16 randomly selected bags have a mean weight that exceeds 15.2 oz is 0.38%.

Z score

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (x - μ)/(σ/√n)

where x is raw score, σ is standard deviation and μ is mean and n is sample size

μ = 15, σ = 0.3, n = 16

For x > 15.2:

z = (15.2 - 15)/(0.3/√16) = 2.67

P(z > 2.67) = 1 - P(z < 2.67) = 1 - 0.9962 = 0.0038

The probability that 16 randomly selected bags have a mean weight that exceeds 15.2 oz is 0.38%.

Find out more on Z score at: https://brainly.com/question/25638875