PRACT STAT W/ ACCESS 6MO LOOSELEAF
PRACT STAT W/ ACCESS 6MO LOOSELEAF
4th Edition
ISBN: 9781319215361
Author: BALDI
Publisher: Macmillan Higher Education
Question
Book Icon
Chapter 16, Problem 16.26AT

(a)

To determine

To find out what is the probability that any one seed weighs more than 525 mg and also between 500 and 550 mg.

(a)

Expert Solution
Check Mark

Answer to Problem 16.26AT

The probability that any one seed weighs more than 525 mg is 0.50 and forseed weighs between 500 and 550 mg is 0.1820 .

Explanation of Solution

In the question, it is given that the distribution of seed weights for a variety of winged bean. This distribution is approximately Normal with,

  μ=525σ=110

And we are considering selecting at random four seeds of this variety. Thus, the probability that any one seed weighs more than 525 mg is calculated as:

  P(X>525)=P(Z>525525110)=P(Z>0)=1P(Z<0)=10.50=0.50

And the probability that any one seed weighs between 500 and 550 mg is calculated as:

  P(500<X<550)=P(500525110<Z<550525110)=P(0.23<Z<0.23)=P(Z<0.23)P(Z<0.23)=0.59100.4090=0.1820

(b)

To determine

To find out what is the probability that all four seeds in the sample weighs more than 525 mg and also find out what is the probability that the average weight of the four seeds is greater than 525 mg and which of these two probabilities is the smallest.

(b)

Expert Solution
Check Mark

Answer to Problem 16.26AT

The probability that all four seeds in the sample weighs more than 525 mg is 0.0625 and the probability that the average weight of the four seeds is greater than 525 mgis 0.50 .

Explanation of Solution

In the question, it is given that the distribution of seed weights for a variety of winged bean. This distribution is approximately Normal with,

  μ=525σ=110

And we are considering selecting at random four seeds of this variety. Thus, theprobability that all four seeds in the sample weighs more than 525 mg is calculated as:

   P(X=4)= (0.5) 4 =0.0625

And the probability that the average weight of the four seeds is greater than 525 mg is calculated as:

  P(x¯μσn>5255251104)=P(Z>0)=10.50=0.50

By looking at the above two probabilities we can see that the probability that all four seeds in the sample weighs more than 525 mg is smaller than the probability that the average weight of the four seeds is greater than 525 mg because observing four seeds that are all over 525 mg would be difficult if each individual seed only have 50% chance of weighing more than 525 mg.

(c)

To determine

To find out what is the probability that all four seeds in the sample weighs between 500 and 550 mg and also find out what is the probability that the average weight of the four seeds is between 500 and 550 mg and how does this relate to your explanation for part (b).

(c)

Expert Solution
Check Mark

Answer to Problem 16.26AT

The probability that all four seeds in the sample weighs between 500 and 550 mg is 0.0011 and the probability that the average weight of the four seeds is between 500 and 550 mgis 0.3472 .

Explanation of Solution

In the question, it is given that the distribution of seed weights for a variety of winged bean. This distribution is approximately Normal with,

  μ=525σ=110

And we are considering selecting at random four seeds of this variety. Thus, the probability that all four seeds in the sample weighs between 500 and 550 mgis calculated as:

   P(X=4)= (0.1820) 4 =0.0011

And the probability that the average weight of the four seeds is between 500 and 550 mgis calculated as:

  P(5005251104<x¯μσn<5505251104)=P(0.45<Z<0.45)=P(Z<0.45)P(Z<0.45)=0.67360.3264=0.3472

Both of these probabilities are smaller relative to the corresponding answer from part (b). Since the probability an individual seed weighs between 500 and 550 mgis smaller than the probability an individual seed weighs over 525 mg, selecting four seeds that are all between 500 and 550 mg or observing an average of four seeds between 500 and 550 mg will be even more difficult.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Suppose the Internal Revenue Service reported that the mean tax refund for the year 2022 was $3401. Assume the standard deviation is $82.5 and that the amounts refunded follow a normal probability distribution. Solve the following three parts? (For the answer to question 14, 15, and 16, start with making a bell curve. Identify on the bell curve where is mean, X, and area(s) to be determined. 1.What percent of the refunds are more than $3,500? 2. What percent of the refunds are more than $3500 but less than $3579? 3. What percent of the refunds are more than $3325 but less than $3579?
A normal distribution has a mean of 50 and a standard deviation of 4. Solve the following three parts? 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the answer of the second part.) 2. Compute the probability of a value greater than 55.0. Use the same formula, x=55 and subtract the answer from 1. 3. Compute the probability of a value between 52.0 and 55.0. (The question requires finding probability value between 52 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 52, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…
If a uniform distribution is defined over the interval from 6 to 10, then answer the followings: What is the mean of this uniform distribution? Show that the probability of any value between 6 and 10 is equal to 1.0 Find the probability of a value more than 7. Find the probability of a value between 7 and 9.   The closing price of Schnur Sporting Goods Inc. common stock is uniformly distributed between $20 and $30 per share. What is the probability that the stock price will be: More than $27? Less than or equal to $24?   The April rainfall in Flagstaff, Arizona, follows a uniform distribution between 0.5 and 3.00 inches. What is the mean amount of rainfall for the month? What is the probability of less than an inch of rain for the month? What is the probability of exactly 1.00 inch of rain? What is the probability of more than 1.50 inches of rain for the month? The best way to solve this problem is begin by a step by step creating a chart. Clearly mark the range, identifying the…
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman