An athletic director found that 31% of the football players have an A average in school. If 2% of the students at the school play football, what is the probability that a student chosen at random will be a football player with an A average?

Question

An athletic director found that 31% of the football players have an A average in school. If 2% of the students at the school play football, what is the probability that a student chosen at random will be a football player with an A average?

suppose a normal distribution has a true population mean of 18 and a true standard deviation of 2. What % of observations will be 18 or less?

Answer 1

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

In a normal distribution, mean = mode = median. What is the definition of median?

Answer 2

To find the probability that a student chosen at random will be a football player with an A average, we need to use conditional probability.

Let's break down the information given:

- The percentage of football players with an A average is 31% = 0.31.
- The percentage of students who play football is 2% = 0.02.

To find the probability, we can multiply the probability of being a football player by the probability of having an A average:

Probability (football player with an A average) = Probability (football player) * Probability (A average)

= 0.02 * 0.31

= 0.0062 (or 0.62%)

So, the probability that a student chosen at random will be a football player with an A average is 0.62%.

Now, let's move on to the second question:

To find the percentage of observations that are 18 or less in a normal distribution with a mean of 18 and a standard deviation of 2, we can use the z-score formula and the standard normal distribution table.

The z-score formula is calculated by subtracting the mean from the observation and dividing it by the standard deviation:

z = (observation - mean) / standard deviation

In this case, we want to find the percentage of observations that are 18 or less, which means we are interested in the area to the left of 18 on the normal distribution curve.

To calculate this, we need to find the z-score for 18, which can be calculated as:

z = (18 - 18) / 2

z = 0 / 2

z = 0

The z-score of 0 corresponds to the mean value in the standard normal distribution.

Since the mean of the standard normal distribution is 0 and the area under the curve to the left of 0 is 0.5, we can conclude that:

The percentage of observations that are 18 or less in a normal distribution with a mean of 18 and a standard deviation of 2 is 50%.