Respuesta :
Let [tex]X[/tex] be the random variable for the value of goods sold in any given month. Also let [tex]Y[/tex] be the random variable representing the number of times that sales in a given month surpass a value of 100.
[tex]X[/tex] follows a normal distribution with mean [tex]\mu=100[/tex] and standard deviation [tex]\sigma=5[/tex], while [tex]Y[/tex] would follow a binomial distribution with some success probability [tex]p[/tex] on [tex]n=6[/tex] trials. To find [tex]p[/tex], we use the distribution of [tex]X[/tex]:
[tex]p=P(X>100)=P\left(\dfrac{X-100}5>\dfrac{100-100}5\right)=P(Z>0)=\dfrac12[/tex]
where [tex]Z[/tex] is a random variable following the standard normal distribution.
a) The probability that exactly 3 of the next 6 months have sales greater than 100 is then
[tex]P(Y=3)=\dbinom63\left(\dfrac12\right)^3\left(1-\dfrac12\right)^{6-3}=\dfrac{6!}{3!(6-3)!}\cdot\dfrac1{2^6}=\dfrac5{16}\approx0.31[/tex]
b) Let [tex]W[/tex] be the random variable representing the total value of sales over 4 months, and denote by [tex]X_i[/tex] the random variable for the value of sales during month [tex]i[/tex]. Then [tex]W=X_1+X_2+X_3+X_4[/tex], and we want to find [tex]P(W>420)[/tex].
We know the [tex]X_i[/tex] are mutually independent and identically distributed, and we (you probably should, anyway) also know that the sum of i.i.d. normally distributed random variables also follows a normal distribution, whose mean is the sum of the means, and whose standard deviation is the sum of the squares of the standard deviations, of the i.i.d. random variables:
[tex]X_i\sim\mathcal N(100,5)\implies W\sim\mathcal N(400,100)[/tex]
Then the probability we want is
[tex]P(W>420)=P\left(\dfrac{W-400}{100}>\dfrac{420-400}{100}\right)=P(Z>0.2)\approx0.42[/tex]
The probability that exactly 3 of the next 6 months have sales greater than 100 is 0.31 and the probability that the total of the sales in the next 4 months is greater than 420 is 0.42.
Given :
Monthly sales are independent normal random variables with a mean of 100 and a standard deviation of 5.
The probability is given by the formula:
[tex]\rm p = P(X>100) = P\left(\dfrac{X-100}{5 }>\dfrac{100-100}{5}\right)=P(Z>0)=\dfrac{1}{2}[/tex]
a) The probability that exactly 3 of the next 6 months have sales greater than 100 is given by:
[tex]\rm P(Y=3)=\dfrac{6}{3}\times \left(\dfrac{1}{2}\right)^2\times \left(1-\dfrac{1}{2}\right)^{6-3}[/tex]
[tex]\rm P(Y=3)=\dfrac{5}{16}\approx 0.31[/tex]
b) The probability that the total of the sales in the next 4 months is greater than 420 is given by:
[tex]\rm P(W>420) = P\left(\dfrac{W-400}{100 }>\dfrac{420-400}{100}\right)=P(Z>0.2)\approx 0.42[/tex]
For more information, refer to the link given below:
https://brainly.com/question/23044118
