Respuesta :
Answer:
[tex] z = \frac{14.3-22.4}{2.7}=-3[/tex]
Since we know that 99.7% of the data are between 3 deviations from the mean we can conclude that:
[tex] P(X <14.3)= \frac{1-0.997}{2}= 0.0015[/tex]
Since we know that the normal distribution is symmetric for this reason we divide by 2
Step-by-step explanation:
For this case we know that the lifespans of tigers X are normally distributed:
[tex] X \sim N (\mu = 22.4 , \sigma =2.7)[/tex]
From the empirical rule we know that we have 68% of the values within one deviation from the mean, 95% within 2 deviations and 99.7% within 3 deviations.
We want to find this probability:
[tex] P(X<14.3)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
In order to find how many deviations above/below we are from the mean. And if we find the z score for 14.3 we got:
[tex] z = \frac{14.3-22.4}{2.7}=-3[/tex]
Since we know that 99.7% of the data are between 3 deviations from the mean we can conclude that:
[tex] P(X <14.3 )= \frac{1-0.997}{2}= 0.0015[/tex]
Since we know that the normal distribution is symmetric for this reason we divide by 2
