Mathematics, 10.03.2020 00:47 arunamvr

Suppose that 30% of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other 70% want a used copy. Consider randomly selecting 25 purchasers. a. What are the mean value and standard deviation of the number who want a new copy of the book? b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value?

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Answer from: vlactawhalm29

Mean = 7.5 and standard deviation = 2.29, b) 0.017.fgdgsgdg

Step-by-step explanation:

Since we have given that

Probability of all students who have to buy a text for a particular = 30%

Probability of all students who want a used copy = 70%

Number of purchasers = 25

a. What are the mean value and standard deviation of the number who want a new copy of the book?

Mean = E[x] = $np=25\times 0.3=7.5$

Standard deviation = SD[x] = $\sqrt{npq}=\sqrt{25\times 0.3\times 0.7}=2.29$

b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value?

$P(X\mu+2\sigma)=P(X7.5+2\times 2.29)=P(X12.08)\\\\=P(X\geq 13)\\\\=1-P(X\leq 12)\\\\=1-\sum _{x=0}^{12}^{25}C_x(0.3)^x(0.7)^{25-x}\\\\=1-0.983\\\\=0.017$

Hence,a) mean = 7.5 and standard deviation = 2.29, b) 0.017.

Answer from: jade611135

a) Mean = 7.5

Standard deviation = 2.29

b) 0.0174

Step-by-step explanation:

We are given the following information:

We treat student who have to buy a new text as a success.

P(student who have to buy a new copy) = 30% = 0.3

p = 0.3

P(student who have to buy a used copy) = 70% = 0.7

Sample size, n = 25

Then the number of adults follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

a) mean value and standard deviation of the number who want a new copy

$\mu = np = 25(0.3) = 7.5\\\sigma = \sqrt{np(1-p)} = \sqrt{25(0.3)(0.7)} = 2.29$

b) P(number who want new copies is more than two standard deviations away from the mean value)

$P(x \mu + 2\sigma)\\=P(x 7.5 + 2(2.29)\\=P(x 12.08)\\=P(x13)\\=1-P(x \leq 12) = 1-(P(x = 0) +...+ P(x = 12)) \\=1- (\binom{25}{0}(0.30)^0(1-0.30)^{25} +...+ \binom{25}{12}(0.30)^{12}(1-0.30)^{13})\\=1-0.9825\\= 0.0174$

0.0174 is the probability that the number who want new copies is more than two standard deviations away from the mean value.

Answer from: Quest

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

Answer from: Quest

You need to divide 210 square inches by 35 inches and that’s going to give you 6 inches and that’s the length of the hypothenuse of the right triangle i hope that this you : )
Pllzz hellp ! will give branliest !

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