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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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.

2. (30) 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:
А. Р(X %3D 3).
В. Р(X 3 6).
С. Р(X < 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-

$P (X = x) = \frac{{}^{N_{1}}C_{x} \times {}^{N_{2}}C_{n - x}}{{}^{N}C_{n}}, w h e r e N = N_{1} + N_{2}$

Therefore,

$\begin{array}{rcl} P (X = 3) & = & \frac{{}^{11}C_{3} \times {}^{14}C_{(7 - 3)}}{{}^{25}C_{7}} \\ = & \frac{\frac{11!}{3! \times 8!} \times \frac{14!}{4! \times 10!}}{\frac{25!}{7! \times 18!}} \\ = & 0.3436 \end{array}$

$\begin{array}{rcl} P (X = 6) & = & \frac{{}^{11}C_{6} \times {}^{14}C_{7 - 6}}{{}^{25}C_{7}} \\ = & \frac{\frac{11!}{6! \times 5!} \times \frac{14!}{1! \times 13!}}{\frac{25!}{7! \times 18!}} \\ = & 0.0135 \end{array}$

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