Answer to Question #276184 in Statistics and Probability for lan

Question

Factory produces whiteboard markers. The production process can be modelling by normal distribution with a mean length of 11 cm and a standard deviation of 0.25 cm.

a) What is the probability that a randomly selected whiteboard marker has a length longer than 10.5 cm? (2 marks)

100 whiteboard markers are randomly selected for quality checking.

(b) What are the mean and standard deviation of the sample mean length? (2 marks)

(c) What is the probability that the sample mean length will be between 10.99 cm and 11.01 cm? (2 marks)

(d) If 99.8% of the sample means are more than a specific length Y, find Y. (3 marks)

Answer 1

(a) $X\sim N(11, 0.25^2)$

$$P(X>10.5)=1-P(X\leq 10.5)$$

$$=1-P(X\leq\dfrac{10.5-11}{0.25})=1-P(X\leq -2)$$

$$\approx0.97725$$

(b)

$$\mu_{\bar{X}}=\mu_X=11,$$

$$\sigma_{\bar{X}}=\sigma_X/\sqrt{n}=0.25/\sqrt{100}=0.025$$

$\bar{X}\sim N(11, 0.025^2)$

(c)

$$P(10.99<\bar{X}<11.01)$$

$$=P(\bar{X}<11.01)-P(\bar{X}\leq10.99)$$

$$=P(\bar{X}<\dfrac{11.01-11}{0.025})-P(\bar{X}\leq\dfrac{10.99-11}{0.025})$$

$$=P(\bar{X}<0.4)-P(\bar{X}\leq-0.4)$$

$$=0.6554217-0.3445783$$

$$\approx0.310843$$

(d)

$$P(\bar{X}>Y)=0.998$$

$$P(Z\leq\dfrac{Y-11}{0.025})=1-0.998$$

$$\dfrac{Y-11}{0.025}\approx-2.878162$$

$$Y=11-0.025(2.878162)$$ $$Y=10.928$$