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Probability calculation is a crucial topic in statistics, especially when dealing with normal distributions. In this exercise, we're interested in finding the probability that Ken will use more than 0.85 ounces of toothpaste during his trip. This involves assessing whether the total amount of toothpaste used exceeds a given threshold.

In probability terms, we express this as finding the value of \( P(X > 0.85) \), where \( X \) is the normally distributed quantity representing total toothpaste use.

To calculate this, we first standardize \( X \) into a Z-score, which allows us to use standard normal distribution tables. Finding this probability gives insight into how likely it is for Ken to run out of toothpaste, which in our solution was found to be about 0.076, or 7.6%.

This tells us that there's a relatively low chance that Ken will use more than the allotted 0.85 ounces of toothpaste.

The standard deviation is a measure of how much variation or dispersion exists from the average (mean) of a set of values. In the context of this problem, the standard deviation tells us how much Ken's toothpaste usage deviates from the average amount he squeezes out each time he brushes his teeth.

For the amount of toothpaste per brushing:

• The population standard deviation \( \sigma \) is given as 0.02 ounces.
• When Ken brushes six times, we need to adjust this to reflect the variability in total usage.

The total standard deviation for six brushes is calculated using the formula \( \sigma_X = \sqrt{n} \times \sigma \), where \( n = 6 \).

Therefore, \( \sigma_X = \sqrt{6} \times 0.02 \approx 0.04899 \) ounces." This adjusted standard deviation shows the spread of tooth paste usage when considering all six brushing episodes together.

A Z-score, also known as a standard score, quantifies how many standard deviations an element is from the mean. In the context of normal distribution, the Z-score helps to transform any normal random variable into the standard normal distribution (mean 0, standard deviation 1), making it easier to determine probabilities.

To find the Z-score for the total toothpaste usage (\( X \)), we use the formula: \[ Z = \frac{X - \mu_X}{\sigma_X} \]

Using our values, \( X = 0.85 \), \( \mu_X = 0.78 \), and \( \sigma_X = 0.04899 \), the calculation becomes: \[ Z = \frac{0.85 - 0.78}{0.04899} \approx 1.429 \]

This Z-score allows us to determine the probability of Ken using more than 0.85 ounces from standard normal distribution tables. This score indicates how unusual or typical an outcome is, given the distribution parameters.

The mean of a dataset is a measure of its central tendency, often referred to as the "average." In this situation, the mean provides the average amount of toothpaste Ken uses in a single brushing session, which is crucial for calculating the total expected use over multiple sessions.

Given that Ken uses an average of \( 0.13 \) ounces per brushing, when considering all six brushing events, the total mean (\( \mu_X \)) is calculated by multiplying the mean per brushing by the number of brushings: \[ \mu_X = 6 \times 0.13 = 0.78 \text{ ounces} \]

This total mean represents the expected toothpaste usage over the course of the six brushing sessions during his trip. Understanding the mean helps set a baseline expectation and serves as the cornerstone for calculating probabilities and assessing variations in Ken's usage pattern.