Answer to Question #167088 in Statistics and Probability for ira

Question

the time taken by students to solve a computer assignment produces a normal distribution with a mean of 33.0 minutes and standard deviation of 5.0 minutes. if a random sample is selected, find the probability that the student will solve:

a) in 19.1 to 28.2 minutes

b) less than 27.5 minutes

c) more than 35 minutes

Answer 1

Let $X=$ the time taken by students to solve a computer assignment: $X\sim N(\mu, \sigma^2)$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

Given $\mu=33.0\ min, \sigma=5.0\ min.$

a)

$$P(19.1<X<28.2)=P(X<28.2)-P(X\leq19.1)$$

$$=P(Z<\dfrac{28.2-33}{5})-P(Z\leq\dfrac{19.1-33}{5})$$

$$=P(Z<-0.96)-P(Z\leq-2.78)$$

$$\approx0.16853-0.00272\approx0.1658$$

b)

$$P(X<27.5)=P(Z<\dfrac{27.5-33}{5})=P(Z<-1.1)$$

$$\approx0.1357$$

c)

$$P(X>35)=1-P(X\leq35)$$

$$=1-P(Z\leq\dfrac{35-33}{5})=1-P(Z\leq0.4)$$

$$\approx1-0.65542\approx0.3446$$