the average number of passengers on a subway car is 60. If the number of passengers on a car has normal distribution with a standard deviation of 20, approximately what the percent of subway cars carry more than 80 passengers?

Question

the average number of passengers on a subway car is 60. If the number of passengers on a car has normal distribution with a standard deviation of 20, approximately what the percent of subway cars carry more than 80 passengers?

Answer 1

your data is perfectly suited for this webpage, which replaces the tables you probably have at the back of your text

I got 15.87%

http://davidmlane.com/hyperstat/z_table.html

Answer 2

To find the percentage of subway cars that carry more than 80 passengers, we first need to standardize the value of 80 using the given mean and standard deviation.

First, we calculate the z-score using the formula:

z = (x - μ) / σ

Where:
x = value we want to standardize (80)
μ = mean (average number of passengers on a subway car) = 60
σ = standard deviation = 20

Substituting the values into the formula:

z = (80 - 60) / 20
z = 20 / 20
z = 1

Now, we need to find the area to the right of the z-score of 1 in the standard normal distribution table.

Consulting the z-table, we find that the area to the right of the z-score of 1 is approximately 0.1587.

This means that about 15.87% of subway cars carry more than 80 passengers.

Answer 3

To find the percent of subway cars that carry more than 80 passengers, we need to first calculate the z-score of 80 in relation to the normal distribution.

The formula to calculate the z-score is:

z = (x - μ) / σ

Where:
- x is the value we want to standardize (in this case, 80)
- μ is the mean (average) of the distribution (in this case, 60)
- σ is the standard deviation of the distribution (in this case, 20)

Plugging in the values, we get:

z = (80 - 60) / 20
z = 20 / 20
z = 1

The z-score tells us how many standard deviations 80 is from the mean.

Next, we need to find the percentage of values that fall above this z-score using a standard normal distribution table (also known as a z-table). The z-table gives us the cumulative probability up to a certain z-score.

Looking up a z-score of 1 in the z-table, we find that the cumulative probability is 0.8413. This means that 84.13% of subway cars have less than or equal to 80 passengers.

To find the percentage of subway cars that carry more than 80 passengers, subtract this cumulative probability from 1:

1 - 0.8413 = 0.1587

So, approximately 15.87% of subway cars carry more than 80 passengers.