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

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

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