Question

A local casino provides bus service for gamblers, so they can get to the casino. Each bus holds 55 people and research shows that individuals have a mean loss of $182 with a standard deviation of $52.50 when gambling. If gambling loses are normally distributed then how many people on a full bus do, we expect will lose at least $200?3. A local casino provides bus service for gamblers, so they can get to the casino. Each bus holds 55 people and research shows that individuals have a mean loss of $182 with a standard deviation of $52.50 when gambling. If gambling loses are normally distributed then how many people on a full bus do, we expect will lose at least $200?

Answer 1

To estimate the number of people expected to lose at least $200, calculate the z-score for a $200 loss, look up the corresponding probability in the standard normal distribution table, subtract from 1, and multiply by the number of people on the bus (55).

Step-by-step explanation:

To find out how many people on a full bus we expect will lose at least $200, we need to use the normal distribution properties given the mean loss of $182 and a standard deviation of $52.50.

First, we calculate the z-score for a loss of $200:

Z = (X - μ) / σ

Z = ($200 - $182) / $52.50

Z = $18 / $52.50

Z = 0.34286

Now we consult the standard normal distribution table to find the probability that X is greater than or equal to $200, which corresponds to the z-score we calculated. Since the normal distribution table gives the probability that Z is less than a given value, we subtract this value from 1 to get the probability that Z is greater than 0.34286.

If the value from the table is represented as P(Z < 0.34286), then the probability that a person will lose at least $200 is 1 - P(Z < 0.34286). Let's say this value is P1. We then multiply this probability by the number of people on the bus (55) to find the expected number of people:

Expected number of people losing at least $200 = 55 * P1