Question

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears filled with blackboard errors, the total amount in dollarspaid can be approximated by a Gaussian random variableYnwith expected value 40nandand variance 100n. What is the probability that 20exceeds 1000

Answer 1

The correct answer is "0.0000039110".

Explanation:

The given values are:

`Y_n\rightarrow N(\mu, \sigma^2)`

`\mu = 40n`

`\sigma^2=100n`

`n=20`

then,

The required probability will be:

=
`P(Y_(20)>1000)`

=
`P((Y_(20)-\mu)/(\sigma) >(1000-40* 20)/(√(100* 20) ) )`

=
`P(Z>(1000-800)/(44.7214) )`

=
`P(Z>(200)/(44.7214) )`

=
`P(Z>4.47)`

By using the table, we get

=
`0.0000039110`