Prob and Stat Homework Help

Question

Suppose certain coins have weights that are normally distributed with a mean of 5.493 g

and a standard deviation of 0.051 g. A vending machine is configured to accept those coins with weights between 5.413 g and 5.573 g.

a. If 290 different coins are inserted into the vending​ machine, what is the expected number of rejected​ coins?

The expected number of rejected coins is:

Answer 1

Hi! So it sounds like the coin weight (let's call it W) is normally distributed with mean 5.493 g and standard deviation of 0.051 grams. We want to find the probability that the weight of the coin is either less than 5.413 g or greater than 5.573 g.

We'll use normal approximation here to get the probabilities, which means we'll need z-scores:

P(W<5.413). To get the z-score corresponding to 5.413, we subtract the mean and then divide by the standard deviation:

z= (5.413-5.493)/0.051 = -1.569

We can either look up this value on a z-table or use the R command pnorm(-1.569) to find that

P(z < -1.569) = 0.0583

P(W>5.573): Whenever you're using z-scores. you want to make sure everything's written in terms of a less than, so we're instead gonna write this as 1-P(W<5.573). Again, we'll get the z-score by subtracting the mean and dividing by the standard deviation.

z = (5.573-5.493)/0.052 = 1.53849

Again, we can either use a z table or the pnorm command:

pnorm(1.1538)= 0.8757

1 - .8757 = 0.1243

Add these two probabilities together to get:

0.0583 + 0.1243 = 0.1826

and that's your answer :)