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?

Short answer

The standard deviation needs to be approximately 0.343 ounces.

Step-by-step solution

Understanding the Problem

We have a normal distribution with mean 12.8 ounces and need to adjust the standard deviation so only 1% of the distribution is less than 12 ounces. We aim to find the new standard deviation.

Identify the Z-score for 1%

In a standard normal distribution, 1% in the lower tail corresponds to a Z-score. We look up this Z-score using a standard normal distribution table or calculator. The Z-score for 1% in the lower tail is approximately -2.33.

Set Up the Equation

We use the formula for a normal distribution: \[x = \mu + z\sigma\]where \(x = 12\), \(\mu = 12.8\), \(z = -2.33\), and \(\sigma\) is the unknown standard deviation. Set up the equation: \[12 = 12.8 + (-2.33)\sigma\]

Solve for the New Standard Deviation

Rearrange the equation to solve for \(\sigma\):\[12 - 12.8 = -2.33\sigma \]\[-0.8 = -2.33\sigma\]\[\sigma = \frac{-0.8}{-2.33} \approx 0.343\]

Key concepts

Standard Deviation

The standard deviation is a measure of how spread out the values are in a data set. Think of it like a measure of how much the data "deviates" from the average or mean. A smaller standard deviation indicates that the data points are close to the mean. A larger standard deviation means the data can be very spread out.
When dealing with a normal distribution, the standard deviation helps determine how the data is dispersed around a mean value. For example, in our potato chip exercise, the standard deviation tells us how the weights of the bags cluster around the mean of 12.8 ounces.

• A standard deviation of 0.6 ounces means that the weights are fairly varied around the mean.
• Adjusting the standard deviation to 0.343 ounces makes the weights less varied, ensuring that fewer bags fall below 12 ounces.

The relationship between the mean, the standard deviation, and individual data points is crucial for understanding how often a certain measurement, like the weight of a chip bag, will occur.

Z-Score

The Z-score measures how many standard deviations a data point is from the mean. In other words, it describes a data point's position relative to the mean and the spread (measured by the standard deviation) of the entire data set.
To calculate a Z-score, we use the formula: \[z = \frac{x - \mu}{\sigma}\] where \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
In our potato chip example, we want to find a Z-score representing only 1% in the lower tail of a standard normal distribution, which corresponds to approximately -2.33. This Z-score helps us determine how the weights under this point are distributed.

• A negative Z-score indicates the value is below the mean, like the 12 ounces we looked at in the exercise.
• By solving for the new standard deviation with the Z-score, we can tailor the distribution to put only 1% below a certain threshold.

This understanding allows manufacturers to adjust the standard deviation to control quality successfully.

Mean

The mean, often referred to as the average, is the sum of all values in the data set divided by the number of values. It serves as a "central" value for the data.
In a normal distribution, data tends to cluster around the mean, creating the bell curve shape typical of such distributions.

• In our context, the mean weight of the potato chip bags is 12.8 ounces.
• This mean tells us the main target weight manufacturers aim to achieve.

Maintaining this mean while adjusting the standard deviation means they can ensure the bags stay close to the desired weight.
The mean is a foundational concept because it provides us with a starting point to understand the overall distribution and behavior of a dataset. By examining how other statistics, like the standard deviation and Z-score, relate to the mean, we can gain valuable insights into the data's overall pattern and how to manage it effectively.