Mathematics, 28.06.2019 13:30 mitch729

Monthly sales are independent normal random variables with mean 100 and standard deviation 5. (a) find the probability that exactly 3 of the next 6 months have sales greater than 100. (b) find the probability that the total of the sales in the next 4 months is greater than 420

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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:

$\rm p = P(X100) = P\left(\dfrac{X-100}{5 }\dfrac{100-100}{5}\right)=P(Z0)=\dfrac{1}{2}$

a) The probability that exactly 3 of the next 6 months have sales greater than 100 is given by:

$\rm P(Y=3)=\dfrac{6}{3}\times \left(\dfrac{1}{2}\right)^2\times \left(1-\dfrac{1}{2}\right)^{6-3}$

$\rm P(Y=3)=\dfrac{5}{16}\approx 0.31$

b) The probability that the total of the sales in the next 4 months is greater than 420 is given by:

$\rm P(W420) = P\left(\dfrac{W-400}{100 }\dfrac{420-400}{100}\right)=P(Z0.2)\approx 0.42$

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Answer from: kat2788

Let be the random variable for the value of goods sold in any given month. Also let be the random variable representing the number of times that sales in a given month surpass a value of 100.

follows a normal distribution with mean $\mu=100$ and standard deviation $\sigma=5$ , while would follow a binomial distribution with some success probability on n=6 trials. To find , we use the distribution of :

$p=P(X100)=P\left(\dfrac{X-100}5\dfrac{100-100}5\right)=P(Z0)=\dfrac12$

where 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

$P(Y=3)=\dbinom63\left(\dfrac12\right)^3\left(1-\dfrac12\right)^{6-3}=\dfrac{6!}{3!(6-3)!}\cdot\dfrac1{2^6}=\dfrac5{16}\approx0.31$

b) Let be the random variable representing the total value of sales over 4 months, and denote by X_i the random variable for the value of sales during month . Then W=X_1+X_2+X_3+X_4 , and we want to find P(W420) .

We know the X_i 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:

$X_i\sim\mathcal N(100,5)\implies W\sim\mathcal N(400,100)$

Then the probability we want is

$P(W420)=P\left(\dfrac{W-400}{100}\dfrac{420-400}{100}\right)=P(Z0.2)\approx0.42$

Answer from: Quest

$450.00 for 6 weeks

$300

step-by-step explanation:

$200.00 for first week $50.00 x5 equals $250 plus $200 equals $450.00 for six weeks.

$200 for first week

$200 x 0.10 = $20

$20 x 5 = $100

$200 + $100 = $300

Answer from: Quest

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.

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