Question

next may an ornithologist intends to trap one male Ice Creamoo and one female Ice Creamoo. The mass M of the male Ice Creamoo may be regarded as being a normal random variable with mean 116g and standard deviation 16g. The mass F of the female Ice Creamoo may be regarded as being independent of M and as being a normal random variable with mean 106g and standard deviation 12g.
Suppose that one of the two trapped birds escapes. Assuming that the remaining bird will be equally to be the male or the female, determine the probability that it's mass will be more than 118g.

Answer 1

sorry I'm kind of drawing a blank on how to do this

@SemiDefinite @DuartMe @imqwerty when you get a chance, would you mind taking a look at this problem

Answer 2

@DuarteMe

Answer 3

There's a 50-50 chance for the ice cream to be either male or female. We'll have to consider both these possibilities to get the answer.

\(\large{\frac{1}{2}\left(P(Mass > 118 | Male) \right) + \frac{1}{2}\left(P(Mass>188|Female)\right)}\)

Answer 4

since the Mass(M) is a normal random variable we can use the Z-values to get things like \(P(M > some\_value)\)

Answer 5

To use the Z-values table, you'll have to convert the given normal random variable to a standard normal variable.

Answer 6

A Standard normal variable has: mean = 0 and standard deviation = 1

Answer 7

Given a normally distributed variable X with a population mean of \(\mu\) and a population standard deviation of \(\sigma\), \(Z = \large{\frac{X - \mu}{\sigma}}\) would be standard normal.