Question

The amount of time spent by a statistical consultant with a client at their first meeting is a random variable that has a normal distribution with a mean value of 60 minutes and a standard deviation of 10 minutes. a. What is the probability that more than 45 minutes is spent at the first meeting? b. What amount of time is exceeded by only \(10 \%\) of all clients at a first meeting?

Short answer

a. The probability is approximately 0.9332 that more than 45 minutes is spent at the first meeting. b. Only 10% of all clients exceed 72.8 minutes at a first meeting.

Step-by-step solution

Title: Conversion to Z-Score

The first part of the question asks for the probability of the consultant spending more than 45 minutes at the first meeting. The first step to find this is to convert this time to a standard Z-score using the formula: Z = \( \frac{x-\mu}{\sigma} \) where \( x \) = 45 minutes, \( \mu \) = 60 minutes (mean) and \( \sigma \) = 10 minutes (standard deviation). Thus, \( Z = \frac{45 - 60}{10} = -1.5 \)

Title: Probability Calculation

The value of Z-score obtained here means that 45 minutes is 1.5 standard deviation below the mean. Since the question is asking for the probability of spending more than this time, it is actually asking for the area to the right of this Z-score in the standard normal distribution. Using Z-table or software, we find that the area to the right of -1.5 is 0.9332. This value represents the probability.

Title: Conversion using Percentiles

In the second part, we are asked to determine the amount of time that only 10% of the clients exceed. In other words, find the 90th percentile of the amount of time the consultant spends at the first meeting. For a standard normal distribution, the Z value for the 90th percentile can be found using the Z-table or software to be approximately 1.28.

Title: Determine the Time spent

To find the amount of time that corresponds to this Z-score, we use the inverse of the formula used in Step 1: \( x = \mu + z\sigma \) where \( z \) = 1.28, \( \mu \) = 60 minutes, and \( \sigma \) = 10 minutes. Plugging in these values gives us \( x = 60 + 1.28(10) = 72.8 \) minutes. Thus, only 10% of the clients exceed 72.8 minutes in the first meeting.

Key concepts

Z-score

Understanding the Z-score is crucial when dealing with normal distributions. A Z-score, also known as a standard score, measures how many standard deviations an element is from the mean. In simpler terms, it tells us how far and in what direction, an individual measurement is from the average measurement. The formula to calculate a Z-score is:
\( Z = \frac{x - \mu}{\sigma} \)
where \( x \) is the value we're looking at, \( \mu \) is the mean value, and \( \sigma \) is the standard deviation. In the context of our exercise, a Z-score of -1.5 indicates that 45 minutes spent is 1.5 standard deviations below the mean time of 60 minutes.

Probability Calculation

Probability calculation in the realm of normal distribution often involves finding the area under the curve. This area corresponds to the likelihood of a random variable falling within a specified range. To find the probability of the consultant spending more than a given amount of time, we refer to standard normal distribution tables (Z-table), or use software to calculate the area to the right of the calculated Z-score.
This value, denoted as the cumulative probability, tells us the fraction of all possible values that lie below a certain point. For our example, with a Z-score of -1.5, the cumulative probability to the right is 0.9332 indicating a high likelihood (93.32%) that the meeting will last more than 45 minutes.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values are close to the mean, while a high standard deviation indicates a wide range of values. In the formula for the Z-score, the standard deviation plays a critical role as it standardizes values, allowing us to compare scores from different distributions. In the exercise given, a standard deviation of 10 minutes conveys that the lengths of meetings are varied, but mostly cluster around 10 minutes away from the 60-minute average.

Percentiles

Percentiles are a way to understand the position of a particular value within a data set. The nth percentile indicates that n percent of the data falls below that value. For instance, if you are at the 90th percentile in your exam scores, you did better than 90% of the students. In our example, we were asked to find the 90th percentile of the meeting times. This is equivalent to determining the time that 90% of the meetings do not exceed, which translates to finding the Z-score that marks the 90th percentile and then using the inverse Z-score formula to calculate the actual time in minutes.