if you would please help me solve ALL parts of the question by showing how you solved each part. i am so confused and this is a study guide question so if you show work and solve correctly i can learn and be able to answer on the actual test when we have to take it. i am just so stressed and its overwhelming. please help me!!!! if you solve correctly i will gladly rate you.

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
icon
Related questions
Question

if you would please help me solve ALL parts of the question by showing how you solved each part. i am so confused and this is a study guide question so if you show work and solve correctly i can learn and be able to answer on the actual test when we have to take it. i am just so stressed and its overwhelming. please help me!!!! if you solve correctly i will gladly rate you. 

**Problem Statement:**

A jar contains 25 pieces of candy, of which 11 are yogurt-covered nuts and 14 are yogurt-covered raisins. Let \( X \) equal the number of nuts in a random sample of 7 pieces of candy that are selected without replacement. Find:

A. \( P(X = 3) \).

B. \( P(X = 6) \).

C. \( P(X \leq 3) \).

D. The Mean of \( X \).

E. The Variance of \( X \).
Transcribed Image Text:**Problem Statement:** A jar contains 25 pieces of candy, of which 11 are yogurt-covered nuts and 14 are yogurt-covered raisins. Let \( X \) equal the number of nuts in a random sample of 7 pieces of candy that are selected without replacement. Find: A. \( P(X = 3) \). B. \( P(X = 6) \). C. \( P(X \leq 3) \). D. The Mean of \( X \). E. The Variance of \( X \).
Expert Solution
Step 1

Given,

There are N=25 pieces of candy out of which N1=11 are yogurt-covered nuts and N2=14 are yogurt-covered raisins.

A sample of n=7 pieces of candies are drawn without replacement.

Let X denote the number of nuts in the random sample of candies drawn without replacement.

There X follows hypergeometric distribution and the pmf is given as-

PX=x=CxN1×Cn-xN2CnN, where N=N1+N2

Therefore,

A.

PX=3=C311×C7-314C725=11!3!×8!×14!4!×10!25!7!×18!=0.3436

B.

PX=6=C611×C7-614C725=11!6!×5!×14!1!×13!25!7!×18!=0.0135

 

 

steps

Step by step

Solved in 2 steps

Blurred answer
Knowledge Booster
Inequality
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
Elementary Statistics: Picturing the World (7th E…
Elementary Statistics: Picturing the World (7th E…
Statistics
ISBN:
9780134683416
Author:
Ron Larson, Betsy Farber
Publisher:
PEARSON
The Basic Practice of Statistics
The Basic Practice of Statistics
Statistics
ISBN:
9781319042578
Author:
David S. Moore, William I. Notz, Michael A. Fligner
Publisher:
W. H. Freeman
Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
Statistics
ISBN:
9781319013387
Author:
David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:
W. H. Freeman