Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 222 days and standard deviation sigma equals 15 days.

What is the probability that a random sample of 26 pregnancies has a mean gestation period of 216 days or​ less?

The probability that the mean of a random sample of 26 pregnancies is less than 216 days is approximately

Answer 1

`z = (216-222)/((15)/(√(26)))= -2.040`

And we can find this probability on this way:

`P(z<-2.040)=0.0207`

Explanation:

Let X the random variable that represent the lenghts of the pregnencies of a population, and for this case we know the distribution for X is given by:

`X \sim N(222,15)`

Where
`\mu=222` and
`\sigma=15`

We are interested on this probability

`P(\bar X<216)`

The z score formula is given by:

`z=(x-\mu)/((\sigma)/(√(n)))`

And if we find the z score we got:

`z = (216-222)/((15)/(√(26)))= -2.040`

And we can find this probability on this way:

`P(z<-2.040)=0.0207`