/*! 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 13 The weight of a running shoe is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The weight of a running shoe is normally distributed with a mean of 12 ounces and a standard deviation of 0.5 ounce. a. What is the probability that a shoe weighs more than 13 ounces? b. What must the standard deviation of weight be in order for the company to state that \(99.9 \%\) of its shoes weighs less than 13 ounces? c. If the standard deviation remains at 0.5 ounce, what must the mean weight be for the company to state that \(99.9 \%\) of its shoes weighs less than 13 ounces?

Short Answer

Expert verified
a. 2.28% probability b. Standard deviation should be 0.324 ounces c. Mean weight should be 11.455 ounces

Step by step solution

01

Understand the Problem

We need to use the properties of the normal distribution to find probabilities and calculate new parameters for given conditions related to the weights of running shoes.
02

Calculate Probability for More Than 13 Ounces (Part a)

To find the probability a shoe weighs more than 13 ounces, we use the standard normal distribution. First, find the z-score using the formula: \[ z = \frac{X - \mu}{\sigma} = \frac{13 - 12}{0.5} = 2 \] The z-score of 2 corresponds to the probability to the left in a standard normal distribution table of approximately 0.9772. So the probability of more than 13 ounces is:\[ 1 - 0.9772 = 0.0228 \] or 2.28\%.
03

Find New Standard Deviation (Part b)

To find the new standard deviation so that 99.9\% of shoes weigh less than 13 ounces, identify the z-score associated with 0.999 in a standard normal distribution table, which is approximately 3.09. Use the z-score formula:\[ 3.09 = \frac{13 - 12}{\sigma} \] Solving for \(\sigma\), we get:\[ \sigma = \frac{1}{3.09} \approx 0.324 \text{ ounces} \]
04

Find New Mean Weight (Part c)

To determine the new mean weight when \(\sigma = 0.5\) ounces so that 99.9\% of shoes weigh less than 13 ounces, use the z-score formula again:\[ 3.09 = \frac{13 - \mu}{0.5} \] Solving for \(\mu\), we get:\[ \mu = 13 - (0.5 \times 3.09) = 11.455 \text{ ounces} \]

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.

Probability Calculations
Understanding probability calculations can be crucial when dealing with normally distributed data. In statistics, finding how likely an event will happen involves certain steps, especially with a normal distribution.
First, when you're asked about the probability of a shoe weighing more than a certain amount, you're essentially looking for the area under the tail of a bell curve.
  • You take the known mean (average measure) and standard deviation (how spread out the weights are) to figure this out.
  • Using these, you calculate something called a Z-score. This tells you how far and in what direction your specific measure (like 13 ounces) deviates from the mean.

Once you have the Z-score, you can use a standard normal distribution table or a calculator to find the probability related to that score. For instance, in part a of this exercise, the probability a shoe weighs more than 13 ounces was found using these steps, leading to a result of 2.28%.
Standard Deviation Adjustment
Standard deviation is a measure of variation or dispersion in a set of data values. When you need most weights to fall below a certain threshold, adjusting the standard deviation becomes necessary.
Say, you want 99.9% of shoes to weigh less than a certain amount; you first need to know the z-score relating to 99.9%. This score is approximately 3.09, found in a standard normal distribution table.
  • With this z-score, you rearrange the formula for a standard normal distribution.
  • You solve for what the standard deviation should be, if your mean remains the same, which in this case would be 12 ounces.

In the exercise given, working through the formula revealed that adjusting the standard deviation to approximately 0.324 ounces ensured that 99.9% of the shoes didn't exceed 13 ounces.
Z-score Interpretation
The Z-score helps you understand how a specific data point relates to the mean of a set of data. Specifically, it tells you how many standard deviations away a data point is from the mean. This is valuable when interpreting the probability that certain data appears in your normal distribution.
For instance, if a shoe weighing 13 ounces corresponds to a Z-score of 2, it indicates that it trails 2 standard deviations above the mean of 12 ounces.
  • To use Z-scores effectively, start by calculating it using the formula \( z = \frac{X - \mu}{\sigma} \).
  • This formula takes your specific observation \( X \), subtracts the mean \( \mu \), and divides by the standard deviation \( \sigma \).

In part c of the exercise, if we want 99.9% of weights to be under 13 ounces with a given standard deviation, the Z-score changes how you'd adjust your mean to meet this condition. Calculating this yielded a new mean weight of around 11.455 ounces, ensuring the desired distribution.

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

An article in Atmospheric Chemistry and Physics ["Relationship Between Particulate Matter and Childhood Asthma-Basis of a Future Warning System for Central Phoenix" (2012, Vol. 12, pp. 2479-2490)] reported the use of PM10 (particulate matter \(<10 \mu \mathrm{m}\) diameter) air quality data measured hourly from sensors in Phoenix, Arizona. The 24 -hour (daily) mean PM10 for a centrally located sensor was \(50.9 \mu \mathrm{g} / \mathrm{m}^{3}\) with a standard deviation of \(25.0 .\) Assume that the daily mean of PM10 is normally distributed. a. What is the probability of a daily mean of PM10 greater than \(100 \mu \mathrm{g} / \mathrm{m}^{3} ?\) b. What is the probability of a daily mean of PM10 less than \(25 \mu \mathrm{g} / \mathrm{m}^{3} ?\) c. What daily mean of \(\mathrm{PM} 10\) value is exceeded with probability \(5 \% ?\)

An article in Knee Surgery, Sports Traumatology, Arthroscopy ["Arthroscopic Meniscal Repair with an Absorbable Screw: Results and Surgical Technique" (2005, Vol. \(13,\) pp. \(273-279\) ) \(]\) cited a success rate of more than \(90 \%\) for meniscal tears with a rim width under \(3 \mathrm{~mm}\), but only a \(67 \%\) success rate for tears of \(3-6 \mathrm{~mm}\). If you are unlucky enough to suffer a meniscal tear of under \(3 \mathrm{~mm}\) on your left knee and one of width \(3-6 \mathrm{~mm}\) on your right knee, what is the probability mass function of the number of successful surgeries? Assume that the surgeries are independent.

In \(2002,\) the average height of a woman aged \(20-74\) years was 64 inches, with an increase of approximately 1 inch from 1960 (http://usgovinfo.about.com/od/healthcare). Suppose the height of a woman is normally distributed with a standard deviation of 2 inches. a. What is the probability that a randomly selected woman in this population is between 58 inches and 70 inches? b. What are the quartiles of this distribution? c. Determine the height that is symmetric about the mean that includes \(90 \%\) of this population. d. What is the probability that five women selected at random from this population all exceed 68 inches?

3.5 .5 Determine the cumulative distribution function of a binomial random variable with \(n=3\) and \(p=1 / 4\)

Suppose that the number of customers who enter a store in an hour is a Poisson random variable, and suppose that \(P(X=0)=0.05 .\) Determine the mean and variance of \(X .\)

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.