Respuesta :
Answer: 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] = [tex]np=25\times 0.3=7.5[/tex]
Standard deviation = SD[x] = [tex]\sqrt{npq}=\sqrt{25\times 0.3\times 0.7}=2.29[/tex]
b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value?
[tex]P(X>\mu+2\sigma)=P(X>7.5+2\times 2.29)=P(X>12.08)\\\\=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[/tex]
Hence,a) mean = 7.5 and standard deviation = 2.29, b) 0.017.
Answer:
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
[tex]p = 0.3[/tex]
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
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
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
[tex]\mu = np = 25(0.3) = 7.5\\\sigma = \sqrt{np(1-p)} = \sqrt{25(0.3)(0.7)} = 2.29[/tex]
b) P(number who want new copies is more than two standard deviations away from the mean value)
[tex]P(x > \mu + 2\sigma)\\=P(x > 7.5 + 2(2.29)\\=P(x > 12.08)\\=P(x>13)\\=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[/tex]
0.0174 is the probability that the number who want new copies is more than two standard deviations away from the mean value.
