/*! 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 14 Newborn elephant calves usually ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Newborn elephant calves usually weigh between 200 and 250 pounds—until October 2006, that is. An Asian elephant at the Houston (Texas) Zoo gave birth to a male calf weighing in at a whopping 384 pounds! Mack (like the truck) is believed to be the heaviest elephant calf ever born at a facility accredited by the Association of Zoos and Aquariums. If, indeed, the mean weight for newborn elephant calves is 225 pounds with a standard deviation of 45 pounds, what is the probability of a newborn weighing at least 384 pounds? Assume that the weights of newborn elephants are normally distributed.

Short Answer

Expert verified
The probability of a newborn elephant calf weighing at least 384 pounds is approximately 0.0002.

Step by step solution

01

Define the Problem with Normal Distribution Parameters

We are looking for the probability that a normally distributed random variable (the weight of newborn elephant calves) is at least 384 pounds. We know the mean \( \mu = 225 \) pounds and the standard deviation \( \sigma = 45 \) pounds.
02

Convert to Standard Normal Distribution

To find the probability for this specific case, we first need to convert the weight of 384 pounds into a standard normal variable \( Z \) using the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X = 384 \), \( \mu = 225 \), and \( \sigma = 45 \).
03

Calculate the Z-score

Applying the formula, we calculate the Z-score: \[ Z = \frac{384 - 225}{45} = \frac{159}{45} \approx 3.53 \]
04

Find the Probability using Z-table or Calculator

We consult the standard normal distribution table or use a calculator to find the probability for \( Z < 3.53 \). The Z-table gives us the probability that a value falls below 3.53, which is approximately 0.9998.
05

Calculate Probability for At Least 384 Pounds

Since we need the probability of \( Z \geq 3.53 \), we subtract the Z-table value from 1 (as the total probability is 1): \[ P(Z \geq 3.53) = 1 - 0.9998 = 0.0002 \].

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.

Understanding the Z-score
A Z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. By converting a data point into a Z-score, you can determine how many standard deviations it is away from the mean. This process helps in comparing data from different normal distributions.
In Mack's example, the Z-score helps us understand how unusual his birth weight is compared to average elephant calves. The formula for calculating a Z-score is \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the data point (384 pounds), \( \mu \) is the mean (225 pounds), and \( \sigma \) is the standard deviation (45 pounds).
Calculating Mack's Z-score, we find it to be approximately 3.53, indicating that his weight is 3.53 standard deviations above the mean. This means Mack is significantly heavier than the average calf.
Probability Calculation Essentials
When working with normal distributions, calculating probabilities involves finding the likelihood that a variable falls within a certain range. In Mack's case, we want to know how likely it is for an elephant calf to weigh at least 384 pounds.
To find this probability, we first determine the Z-score which reflects how extreme a value is. Using the Z-score of 3.53 for Mack, we can look this value up in a Z-table or use a statistical calculator to get the probability of a calf weighing less than 384 pounds: 0.9998. However, we're interested in how likely a calf weighs at least 384 pounds, so we subtract this from 1:
\[ P(Z \geq 3.53) = 1 - 0.9998 = 0.0002 \]
This result suggests such a large birth weight is extremely rare.
Defining Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the data points typically differ from the mean of the dataset. A smaller standard deviation indicates that the data points tend to be close to the mean, while a larger standard deviation indicates more spread out data.
In the context of newborn elephant calves, with a mean weight of 225 pounds and a standard deviation of 45 pounds, the standard deviation gives us a sense of the typical variation we might expect around this average weight.
The calculation used to identify Mack's uniqueness considers this deviation, showing that his weight is significantly higher, 3.53 times the standard deviation away, demonstrating how standard deviation serves as a critical tool in understanding distribution characteristics.
What is Standard Normal Distribution?
The standard normal distribution is a special case of the normal distribution where the mean is 0 and the standard deviation is 1. This standardized form allows us to use Z-tables to find probabilities, simplifying statistical analyses.
To convert a normally distributed variable, like the weight of elephant calves, into a standard normal distribution, we calculate its Z-score. This conversion allows us to compare different data points and probabilities within the standard normal framework easily.
In Mack’s scenario, by translating his weight to a Z-score, we used standard normal distribution properties to identify how rare his weight is. It makes working with different scenarios straightforward, allowing us to focus on Z-scores to draw conclusions about probabilities and data points spanning various contexts.

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

In a recent year, Delaware had the highest per capita annual income with \(\$ 51,803\). If \(\sigma=\$ 4850\), what is the probability that a random sample of 34 state residents had a mean income greater than \(\$ 50,000 ?\) Less than \(\$ 48,000 ?\)

To help students improve their reading, a school district decides to implement a reading program. It is to be administered to the bottom 5% of the students in the district, based on the scores on a reading achievement exam. If the average score for the students in the district is 122.6, find the cutoff score that will make a student eligible for the program. The standard deviation is 18. Assume the variable is normally distributed.

The mean number of miles driven per vehicle annually in the United States is 12,494 miles. Choose a randomly selected vehicle, and assume the annual mileage is normally distributed with a standard deviation of 1290 miles. What is the probability that the vehicle was driven more than 15,000 miles? Less than 8000 miles? Would you buy a vehicle if you had been told that it had been driven less than 6000 miles in the past year?

The national average SAT score (for Verbal and Math) is 1028 . If we assume a normal distribution with \(\sigma=92,\) what is the 90th percentile score? What is the probability that a randomly selected score exceeds \(1200 ?\)

An instructor gives a 100-point examination in which the grades are normally distributed. The mean is 60 and the standard deviation is 10. If there are 5% A’s and 5% F’s, 15% B’s and 15% D’s, and 60% C’s, find the scores that divide the distribution into those categories.

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.