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Brush your teeth The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation. Ricardo spends less than one minute brushing his teeth about \(40 \%\) of the time. He spends more than two minutes brushing his teeth \(2 \%\) of the time. Use this information to determine the mean and standard deviation of this distribution.

Short Answer

Expert verified
Mean: 1.11 minutes; Standard Deviation: 0.433 minutes.

Step by step solution

01

Understand the given probabilities

We know that the time Ricardo spends brushing his teeth is normally distributed. Less than 1 minute is spent 40% of the time, and more than 2 minutes is spent 2% of the time.
02

Translate probabilities to z-scores

For the normal distribution, 40% corresponds to a left-tailed probability. Therefore, we look up 0.40 in the z-table to find the z-score for less than 1 minute. Similarly, for more than 2 minutes, we look up 0.98 in the z-table as this is a right-tailed probability for 2%.
03

Find corresponding z-scores

From the standard normal distribution table, the z-score corresponding to 40% probability is approximately -0.2533. For the 98% probability, the z-score is approximately 2.054.
04

Set up equations for mean and standard deviation

Using the z-score formula, we have: \( z = \frac{x - \mu}{\sigma} \).For 1 minute: \( -0.2533 = \frac{1 - \mu}{\sigma} \).For 2 minutes: \( 2.054 = \frac{2 - \mu}{\sigma} \).
05

Solve the system of equations

First, solve for \(\sigma\) by subtracting the equations:\[ 2.054 - (-0.2533) = \frac{2 - \mu}{\sigma} - \frac{1 - \mu}{\sigma} \] \[ 2.3073 = \frac{1}{\sigma} \].Then, \(\sigma = \frac{1}{2.3073}\approx 0.433\).Substitute back to find \(\mu\): \(2.054 = \frac{2 - \mu}{0.433} \), \(2 - \mu = 2.054 \times 0.433\), \(2 - \mu = 0.890\), \(\mu = 2 - 0.890 = 1.11\).
06

Conclude with the calculated values

We have calculated that the mean time Ricardo spends brushing his teeth, \(\mu\), is approximately 1.11 minutes, and the standard deviation, \(\sigma\), is approximately 0.433 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean and Standard Deviation
The mean and standard deviation are two of the most important metrics in a normal distribution. The mean, or average, is a measure of the central tendency of the data. It tells us where the center of the data set is located. In the context of our exercise with Ricardo, we calculated the mean, \(\mu\), to be approximately 1.11 minutes. This means, on average, Ricardo spends around 1.11 minutes brushing his teeth.

The standard deviation, \(\sigma\), on the other hand, measures how spread out the values in the data set are. It gives us an idea of the variability or dispersion from the mean. In Ricardo's case, a standard deviation of approximately 0.433 minutes implies that the time he spends brushing his teeth varies, on average, by 0.433 minutes from the mean of 1.11 minutes.

Why are these measures useful? Because they provide a simple summary of the data's characteristics, allowing us to understand the average behavior and variation in Ricardo's tooth brushing habits.
Z-scores
Z-scores are a statistical measurement that describe a value's relation to the mean of a group of values. They are expressed in terms of standard deviations from the mean. A z-score tells us where a specific data point lies in the distribution. In simpler terms, it signifies how many standard deviations a data point is from the mean.

For instance, in our example of Ricardo brushing his teeth, we found the z-scores for two scenarios: brushing less than 1 minute and brushing more than 2 minutes. The z-score for less than 1 minute (40% probability) was -0.2533, indicating that this time is slightly below average relative to the mean and standard deviation. Meanwhile, for spending over 2 minutes (2% probability), the z-score was 2.054, showing it is significantly above the average time.

Z-scores are essential for translating probabilities in the context of the normal distribution, helping us understand how typical or atypical a specific observation is within the distribution.
Probability Distribution
The probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes. In the case of a normal distribution, the data follows a symmetrical bell-shaped curve, where most observations cluster around the central peak and probabilities for values farther from the mean taper off evenly in both directions.

In Ricardo's example of brushing his teeth, the distribution is normal. We use this characteristic to help find probabilities and z-scores. The 40% probability of brushing less than 1 minute and 2% probability for more than 2 minutes are determined using the properties of the normal distribution.

Knowing the distribution type helps us make statistical inferences and understand the likelihood of different outcomes, which is especially useful for predicting behavior or outcomes based on the data's past trends.
Statistics
Statistics is the study of collecting, analyzing, interpreting, presenting, and organizing data. It's a critical component when it comes to understanding large data sets and making informed decisions. Statistical methods allow us to summarize complex data in a meaningful way.

In our exercise about Ricardo's tooth brushing, various statistical approaches were used. Calculating the mean and standard deviation, determining z-scores, and analyzing probability distribution are all statistical methods. These methods together paint a comprehensive picture of Ricardo's habits, providing insights into typical behavior and variability.

Statistics equips us with tools to process and interpret massive amounts of information quickly and succinctly. It helps simplify complex phenomena into understandable metrics, making it an indispensable part of effective decision-making and data analysis.

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