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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

Respuesta :

Answer:

[tex] z = \frac{216-222}{\frac{15}{\sqrt{26}}}= -2.040[/tex]

And we can find this probability on this way:

[tex]P(z<-2.040)=0.0207[/tex]

Step-by-step 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:

[tex]X \sim N(222,15)[/tex]  

Where [tex]\mu=222[/tex] and [tex]\sigma=15[/tex]

We are interested on this probability

[tex]P(\bar X<216)[/tex]

The z score formula is given by:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And if we find the z score we got:

[tex] z = \frac{216-222}{\frac{15}{\sqrt{26}}}= -2.040[/tex]

And we can find this probability on this way:

[tex]P(z<-2.040)=0.0207[/tex]

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