/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 156 The time it takes a cell to divi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 5 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?

Short Answer

Expert verified
(a) 0.13%, (b) 15.87%, (c) 71.65 minutes.

Step by step solution

01

Convert Time to Standard Normal Variable (Part a)

To find the probability that a cell divides in less than 45 minutes, first convert the time of 45 minutes to a standard normal variable. The formula to convert to a z-score is \(z = \frac{X - \mu}{\sigma}\), where \(X\) is the time in minutes, \(\mu = 60\) minutes is the mean, and \(\sigma = 5\) minutes is the standard deviation. Thus, for 45 minutes, \(z = \frac{45 - 60}{5} = -3\).
02

Calculate Probability from Z-score (Part a)

Using a standard normal distribution table or calculator, find the probability of \(z < -3\). This represents the probability that a cell divides in less than 45 minutes. The probability for \(z < -3\) is approximately 0.0013, which is 0.13%.
03

Convert Time to Standard Normal Variable (Part b)

To find the probability that it takes a cell more than 65 minutes to divide, convert 65 minutes to a z-score using the same formula: \(z = \frac{65 - 60}{5} = 1\).
04

Calculate Probability from Z-score (Part b)

Using the standard normal distribution table, find the probability for \(z > 1\). This is the same as \(1 - P(z < 1)\). The probability \(P(z < 1)\) is approximately 0.8413, so \(P(z > 1) = 1 - 0.8413 = 0.1587\) or 15.87%.
05

Determine Z-score for 99% Completion (Part c)

To find the time by which approximately 99% of cells have completed mitosis, determine the z-score that corresponds to 99% in a standard normal distribution. Using the standard normal distribution table or calculator, a cumulative probability of 0.99 corresponds to a z-score of approximately 2.33.
06

Solve for Time from Z-score (Part c)

Convert the z-score back to a time in minutes using the formula \(X = z \cdot \sigma + \mu\). For \(z = 2.33\), the time \(X = 2.33 \cdot 5 + 60 = 71.65\) minutes. Thus, by approximately 71.65 minutes, 99% of cells have completed mitosis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-Score
The z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is the number of standard deviations a data point is from the mean. The formula to calculate the z-score is:
  • \( z = \frac{X - \mu}{\sigma} \)
Here, \( X \) is the data point, \( \mu \) is the mean of the group, and \( \sigma \) is the standard deviation.
For example, if the mean time for cell division is 60 minutes and the standard deviation is 5 minutes, a cell dividing in 45 minutes would have a z-score of \( \frac{45 - 60}{5} = -3 \).
A z-score of -3 indicates the division time is 3 standard deviations below the mean. Z-scores are useful because they allow us to standardize data, making it easier to compare values across different units or scales.
Probability Calculation
Probability calculation involves determining the likelihood of an event occurring. In a normal distribution, we use z-scores to find probabilities. Once a z-score is determined, it can be used with a standard normal distribution table or calculator to find the probability.
To find the probability that a cell divides in less than a given time, like 45 minutes, we convert it to a z-score and check a z-table. For example, with a z-score of -3, the probability is about 0.13%, representing how rare such a quick division is.
Similarly, for events like cells taking over 65 minutes, calculate the z-score (which in this case is 1) and find the probability of exceeding this z-score using the cumulative probability formula:
  • Probability (\(z > 1\)) = 1 - Probability (\(z < 1\))
The result is about 15.87%, signifying a more common occurrence than the prior situation.
Standard Deviation
Standard deviation is a statistic that measures the dispersion or spread of a data set relative to its 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.
For cell division times, a standard deviation of 5 minutes signifies that most cell division times are within 5 minutes of the mean time of 60 minutes. Mathematically, standard deviation (\(\sigma\)) is calculated using:
  • \( \sigma = \sqrt{\frac{\sum (X_i - \mu)^2}{N}} \)
Where \(X_i\) are the individual data points, \(\mu\) is the mean, and \(N\) is the total number of data points.
Understanding standard deviation helps us grasp the variability and reliability of our data, crucial for making predictions and understanding probability in normal distribution scenarios.
Probability Distribution
Probability distribution provides a model for understanding how probabilities are assigned to outcomes in a dataset. For continuous data, the most common is the normal distribution, often depicted as a bell curve.
This curve is symmetric and centered around the mean, which in the cell division example is 60 minutes. The standard deviation determines the curve's width: smaller deviations create a steep curve, while larger deviations yield a wide one.
Knowing the normal probability distribution helps to predict the likelihood of various outcomes. We can determine what percentage of data falls within certain standard deviations of the mean using probability tables or calculators. For instance, approximately 68% of data falls within one standard deviation of the mean in a normal distribution.
In practical terms, understanding these concepts can be very powerful for anticipating outcomes and analyzing data trends in everyday situations and scientific research alike.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The maximum time to complete a task in a project is 2.5 days. Suppose that the completion time as a proportion of this maximum is a beta random variable with \(\alpha=2\) and \(\beta=\) 3\. What is the probability that the task requires more than two days to complete?

Integration by parts is required. The probability density function for the diameter of a drilled hole in millimeters is \(10 e^{-10(x-5)}\) for \(x>5 \mathrm{~mm}\). Although the target diameter is 5 millimeters, vibrations, tool wear, and other nuisances produce diameters larger than 5 millimeters. (a) Determine the mean and variance of the diameter of the holes. (b) Determine the probability that a diameter exceeds 5.1 millimeters.

Without an automated irrigation system, the height of plants two weeks after germination is normally distributed with a mean of 2.5 centimeters and a standard deviation of 0.5 centimeter. (a) What is the probability that a plant's height is greater than 2.25 centimeters? (b) What is the probability that a plant's height is between 2.0 and 3.0 centimeters? (c) What height is exceeded by \(90 \%\) of the plants?

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch Suppose that the specifications require the dot diameter to be between 0.0014 and 0.0026 inch. If the probability that a do meets specifications is to be \(0.9973,\) what standard deviation is needed?

Let the random variable \(X\) denote a measurement from a manufactured product. Suppose the target value for the measurement is \(m\). For example, \(X\) could denote a dimensional length, and the target might be 10 millimeters. The quality loss of the process producing the product is defined to be the expected value of \(\$ k(X-m)^{2},\) where \(k\) is a constant that relates a deviation from target to a loss measured in dollars. (a) Suppose \(X\) is a continuous random variable with \(E(X)=m\) and \(V(X)=\sigma^{2} .\) What is the quality loss of the process? (b) Suppose \(X\) is a continuous random variable with \(E(X)=\mu\) and \(V(X)=\sigma^{2} .\) What is the quality loss of the process?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.