Question

Suppose that weights of bags of potato chips coming from a factory follow a normal distribution with mean 12.8 ounces and standard deviation .6 ounces. If the manufacturer wants to keep the mean at 12.8 ounces but adjust the standard deviation so that only 1% of the bags weigh less than 12 ounces, how small does he/she need to make that standard deviation

Answer 1

To ensure that only 1% of bags weigh less than 12 ounces, the standard deviation needs to be approximately 0.343 ounces.

Step-by-step explanation:

To find the standard deviation that would result in only 1% of bags weighing less than 12 ounces, we need to find the z-score that corresponds to that percentile. Using a standard normal distribution table, we find that the z-score is approximately -2.33.

Now, we can use the formula for z-score to find the standard deviation:

z = (x - mean) / standard deviation

Substituting the given values, we have:

-2.33 = (12 - 12.8) / standard deviation

Solving for the standard deviation, we get:

standard deviation = (12 - 12.8) / -2.33 ≈ 0.343

Therefore, the manufacturer needs to make the standard deviation approximately 0.343 ounces in order to ensure that only 1% of bags weigh less than 12 ounces.

Answer 2

std dev = 0.625

Step-by-step explanation:

Given that weights of bags of potato chips coming from a factory follow a normal distribution with mean 12.8 ounces and standard deviation .6 ounces.

We must change std deviation such that

P(X<12) =0.01

Mean would remain the same as 12.8 oz

From normal distribution score we can find

10th percentile is -1.28\

i.e. we must have

`12.8-1.28*s = 12\\1.28s = 0.8\\s = 0.625`

Std deviation should be 0.625