If you are not going to write the solution in minitab leave it to someone else =

Question

If you are not going to write the solution in minitab leave it to someone else = V0.3437 The mean number of orders filled correctly in a s standard deviation is 0.5863. The probability that all th or 65.4%. The probability that none of the orders are probability that at least two orders are filled correctly ant. Problems for Section 5.2 chains. Pe- Accuracy" e percent- you go to pendently ilities that orrectly? of orders LEARNING THE BASICS 5.9 Determine the following: a. For n = 4 and b. For n-10 and = 0.40, what is P(X = 9)? c. For n = 10 and = 0.50, what is P(X = 8)? d. For n = 6 and = 0.83, what is P(X = 5)? 12 A recent 18- to 29-year-o from "Tablet an the binomial d -0.12, what is P(X = 0)? six 18-to 29 a. four will o b. all six wil Determine the mean and standard deviation of the variable c. at least fo X in each of the following binomial distributions a. n 4and = 0.10 b, n 4 and = 0.40 d. What are 18- to 29 e. What as 5.13 A s question edge of t cided or order is 5 and -0.80 | c, n d, n = 3 and = 0.50 APPLYING THE CONCEPTS the nrice of a stock between the A.B C

Explanation / Answer

Q1. MiniTab Output

Probability Density Function

Binomial with n = 4 and p = 0.12

x P( X = x )
0 0.599695

Q2.

We don't find the mean, sd in minitab, it calculated in hand using mean , sd formula

a.

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 4 * 0.1
= 0.4
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 4 * 0.1 * 0.9
= 0.36
III.
standard deviation = sqrt( variance ) = sqrt(0.36)
=0.6

b.
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 4 * 0.4
= 1.6
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 4 * 0.4 * 0.6
= 0.96
III.
standard deviation = sqrt( variance ) = sqrt(0.96)
=0.979796

c.
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 5 * 0.8
= 4
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 5 * 0.8 * 0.2
= 0.8
III.
standard deviation = sqrt( variance ) = sqrt(0.8)
=0.894427

d.
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 3 * 0.5
= 1.5
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 3 * 0.5 * 0.5
= 0.75
III.
standard deviation = sqrt( variance ) = sqrt(0.75)
=0.866025

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