Question

A survey of teenagers' allowances showed a normal distribution with mean $20 and a standard deviation of $2. What is the probability that a student's allowance would fall between $18 and $22?

Answer 1

\[Z = \frac{Y-\mu}{\sigma}\]
\[\mu = 20\]
\[\sigma =2\]

Y is both 18 and 22

find Z(18) and Z(22)

then use a z probability table.

Answer 2

The empirical rule also says:
Approximately 68% of the data points lie within the range
\[\pm\ \sigma\ of\ \mu\]

Answer 3

^Following on from your previous question.

Answer 4

@TheresaCO What is 68% as adecimal?

Answer 5

0.68

Answer 6

Yes. P($18~$22) = 0.68

Answer 7

thank you again @kropot72

Answer 8

You're welcome :)