Problem 194 - Applied Statistics and Probability for Engineers

Question

The time it takes a cell to divide (called mitosis) is normally distributed with an average time of one hour and a standard deviation of five minutes. (a) What is the probability that a cell divides in less than 45 minutes? (b) What is the probability that it takes a cell more than 65 minutes to divide? (c) By what time have approximately \(99 \%\) of all cells completed mitosis?

Step-by-Step Solution

Answer

(a) 0.0013, (b) 0.1587, (c) 71.65 minutes.

Step 1: Convert Units for Consistency

The average time for cell division is one hour, which is equivalent to 60 minutes. The standard deviation given is five minutes. We will work with all times in minutes for consistency in calculations.

Step 2: Normalize Given Values Using Z-scores

The process involves converting the values to Z-scores using the formula:\[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the time, \(\mu\) is the mean (60 minutes), and \(\sigma\) is the standard deviation (5 minutes).

Step 3a: Calculate Z-score for Less than 45 Minutes

Substitute 45 minutes into the Z-score formula:\[ Z = \frac{45 - 60}{5} = \frac{-15}{5} = -3 \]Now, use the Z-score table to find the probability associated with \(Z = -3\).

Step 3b: Determine Probability for Less than 45 Minutes

The Z-score of \(-3\) corresponds to a very low probability, close to 0. In a standard normal distribution table, \(P(Z < -3)\) is approximately 0.0013, which is the probability that a cell divides in less than 45 minutes.

Step 4a: Calculate Z-score for More than 65 Minutes

Substitute 65 minutes into the Z-score formula:\[ Z = \frac{65 - 60}{5} = \frac{5}{5} = 1 \]Find the probability for \(Z > 1\) using the Z-score table.

Step 4b: Determine Probability for More than 65 Minutes

The Z-score of 1 corresponds to \(P(Z < 1)\) which is approximately 0.8413. Therefore, \(P(Z > 1) = 1 - 0.8413 = 0.1587\), implying the probability that it takes more than 65 minutes for division is 0.1587.

Step 5: Determine Time for 99% Completion of Mitosis

To find the cutoff time \(t\) for which 99\% of cells complete division, you need the Z-score that corresponds to a cumulative probability of 0.99, which is approximately 2.33 according to the Z-score table. Use the inverse of the Z-score formula:\[ t = Z \cdot \sigma + \mu = 2.33 \cdot 5 + 60 = 71.65 \]Thus, by approximately 71.65 minutes, 99\% of cells have completed mitosis.

Key Concepts

Normal Distribution

A normal distribution is a statistical concept that appears frequently in nature and human behavior, representing how data tends to be distributed. In a normal distribution, the dataset creates a bell-shaped curve, symmetrical about the mean. This principle is used to model many natural phenomena, including the time cells take to divide, known as mitosis.

Certain features define a normal distribution:

• The mean, median, and mode of a normal distribution are equal.
• It's shaped like a bell and is symmetrical around its central point.
• The area under the curve represents the total probability, and it is equal to 1.

Understanding normal distribution helps us calculate the likelihood of different outcomes. In the context of cellular mitosis, knowing that the timing follows a normal distribution allows scientists to make predictions about the duration of this process.

Z-scores

Z-scores are a statistical measurement that describes a value's relation to the mean of a group of values. When you have a normal distribution, Z-scores help identify and compare data points efficiently.

The formula for calculating a Z-score is:

• \[ Z = \frac{X - \mu}{\sigma} \]

where:

• \(X\) is the data point you are evaluating,
• \(\mu\) is the mean of the dataset,
• \(\sigma\) is the standard deviation.

Z-scores tell us how many standard deviations a point is from the mean. A Z-score of 0 indicates the data point is exactly at the mean, while positive and negative Z-scores show deviations above or below the mean.

In our exercise, Z-scores are used to determine the probability of cell division taking less than or more than certain times, aiding in predicting mitosis duration behavior.

Standard Deviation

Standard deviation is an essential statistical concept that quantifies the amount of variation or dispersion in a dataset. It tells us how spread out the values in a dataset are in relation to the mean.

A low standard deviation means that most data points are close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range of values. The formula for standard deviation is often represented as:

• \[ \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(X_i - \mu)^2} \]

This involves:

• \(X_i\) being individual data points,
• \(\mu\) as the mean,
• \(N\) as the number of data points.

Understanding standard deviation in the context of cellular mitosis allows researchers to comprehend the consistency or variability in cell division times, which can identify anomalies or understand biological processes more thoroughly.

Mitosis

Mitosis is a biological process where a single cell divides to produce two identical daughter cells. It's a fundamental part of life that ensures tissue growth, repair, and genetic consistency.

Key stages in mitosis include:

• Prophase: Chromosomes condense and spindle fibers form.
• Metaphase: Chromosomes align in the center of the cell.
• Anaphase: Chromatids are pulled apart to opposite poles.
• Telophase: New nuclear membranes form around each set of chromosomes.
• Cytokinesis: The cell divides into two.

In exercises like the one given where timings of mitosis are analyzed, understanding mitosis is crucial. It allows students and researchers alike to appreciate why such precise measurements and probabilities are used to study biological functions. Such statistical analysis can predict how genetic or environmental factors might influence cell division timing.