Taxi fares are normally distributed with a mean fare of $22.27 and a standard deviation of $2.20.
(a) which should have the greater probability of falling between $21 & $24;
the mean of a random sample of 10 taxi fares or the amount of a single random taxi fare? why?
(b) which should have a greater probability of being over $24-the mean of 10 randomly selected taxi fares or the amount of a single randomly selected taxi fare? why?

Step-by-step explanation:

Given that taxi Fares are normally distributed with a mean fare of $22.27 and a standard deviation of $2.20.

For a random single taxi std deviation is 2.20

But for a sample of size 10, std deviation would be

This would be less than the 2.20

Because std devition is less for sample we get a big z score for the sample than the single.  

As positive values of z increase we find that probability would decrease since normal curve is bell shaped.

So single taxi fare would have higher probability than sample.

B) Here >24.

By the same argument we have z value less for single taxi hence the probability for more than that would be less than that of sample size 10

Simplifying

3y + 13 = 2y + 25

Reorder the terms:

13 + 3y = 2y + 25

Reorder the terms:

13 + 3y = 25 + 2y

Solving

13 + 3y = 25 + 2y

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Add '-2y' to each side of the equation.

13 + 3y + -2y = 25 + 2y + -2y

Combine like terms: 3y + -2y = 1y

13 + 1y = 25 + 2y + -2y

Combine like terms: 2y + -2y = 0

13 + 1y = 25 + 0

13 + 1y = 25

Add '-13' to each side of the equation.

13 + -13 + 1y = 25 + -13

Combine like terms: 13 + -13 = 0

0 + 1y = 25 + -13

1y = 25 + -13

Combine like terms: 25 + -13 = 12

1y = 12

Divide each side by '1'.

y = 12

Simplifying

y = 12

Step-by-step explanation: