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

Answer:

The correct answer is "0.0000039110".

Step-by-step 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:

= $P(Y_{20}>1000)$

= $P(\frac{Y_{20}-\mu}{\sigma} >\frac{1000-40\times 20}{\sqrt{100\times 20} } )$

= $P(Z>\frac{1000-800}{44.7214} )$

= $P(Z>\frac{200}{44.7214} )$

= $P(Z>4.47)$

By using the table, we get

= $0.0000039110$