Answer to Question #198570 in Statistics and Probability for fahmida

Question

3) MRP produces lightbulbs whose life follows a normal distribution, with a mean of 1,200 hours and a standard deviation of 250 hours.  a. If we choose a lightbulb at random, what is the probability that its lifetime will be between 900 and 1,300 hours? b. If we choose a lightbulb at random, what is the probability that its lifetime will be over 1400 hours?

4) Let the random variable X follow a normal distribution with μ = 80 and σ2 = 100.

a. Find the probability that X is greater than 60. b. Find the probability that X is greater than 72 and less than 82. c. Find the probability that X is less than 55.

6) Anticipated consumer demand in a restaurant for free-range steaks next month can be modeled by a normal random variable with mean 1,200 pounds and standard deviation 100 pounds. a. What is the probability that demand will exceed 1,000 pounds? b. What is the probability that demand will be between 1,100 and 1,300 pounds?

Answer 1

3) Let $X=$lifetime, $X\sim N(\mu, \sigma^2)$

Given $\mu=1200\ h, \sigma=250\ h.$

a.

$$P(900<X<1300)=P(X<1300)-P(X\leq 900)$$

$$=P(Z<\dfrac{1300-1200}{250})-P(Z\leq \dfrac{900-1200}{250})$$

$$=P(Z<0.4)-P(Z\leq -1.2)$$

$$\approx0.6554217-0.1150697\approx0.540352$$

b.

$$P(X>1400)=1-P(X\leq 1400)$$

$$=1-P(Z\leq \dfrac{1400-1200}{250})$$

$$=1-P(Z\leq 0.8)\approx0.211855$$

4) Let $X\sim N(\mu, \sigma^2)$

Given $\mu=80, \sigma^2=100.$

a.

$$P(X>60)=1-P(X\leq 60)$$

$$=1-P(Z\leq \dfrac{60-80}{10})$$

$$=1-P(Z\leq-2)\approx0.977250$$

b.

$$P(72<X<82)=P(X<82)-P(X\leq 72)$$

$$=P(Z<\dfrac{82-80}{10})-P(Z\leq \dfrac{72-80}{10})$$

$$=P(Z<0.2)-P(Z\leq -0.8)$$

$$\approx0.5792597-0.2118554\approx0.367404$$

c.

$$P(X<55)=P(Z< \dfrac{55-80}{10})$$

$$=P(Z< -2.5)\approx0.006210$$

6) Let $X=$demand, $X\sim N(\mu, \sigma^2)$

Given $\mu=1200\ pounds, \sigma=100 \ pounds.$

a.

$$P(X>1000)=1-P(X\leq 1000)$$

$$=1-P(Z\leq \dfrac{1000-1200}{100})$$

$$=1-P(Z\leq -2)\approx0.977250$$

b.

$$P(1100<X<1300)=P(X<1300)-P(X\leq 1100)$$

$$=P(Z<\dfrac{1300-1200}{100})-P(Z\leq \dfrac{1100-1200}{100})$$

$$=P(Z<1)-P(Z\leq -1)$$

$$\approx0.8413447-0.1586553\approx0.682689$$