Bodytemperatures ofhealthy koalasarenormally distributed with a mean of 35.6°C and a standard deviation of 1.3°C. a.What is the probability that a health koalahas a body temperature lessthan 35.0°C? (1pt)b.Veterinarians at a nature preserve in Australiathought a population of koalas might be infected with a virus, so they chose a random sample of n=30koalas and measured their body temperatures. If they got a sample mean body temperature less than 35.0°C, do you think that this population of koalas is healthy? Why/why not?Be specific.(4pt)c.List a potential confounding variable for this study and briefly how it might have impactedthe results. (2pt)2.

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was [tex]P(X < 35) = 32.3 \%[/tex] but when the population was taken into account (i.e n = 30) the probability became

[tex]P(\= X < 35) = 0.6 \%[/tex]

Step-by-step explanation:

From the question we are told that

The mean is [tex]\mu = 35.6^oC[/tex]

The standard deviation is [tex]s = 1.3^oC[/tex]

The sample size is n = 30

Generally the probability that a health koala has a body temperature less than 35.0°C is mathematically represented as

[tex]P(X < 35) = P(\frac{X - \mu }{s} < \frac{35 - 35.6}{1.3} )[/tex]

Here [tex](\frac{X - \mu }{s} = Z (The \ standardized \ value \ of \ X )[/tex]

So

[tex]P(X < 35) = P(Z < -0.46)[/tex]

From the z-table P(Z < -0.46) = 0.323

So

[tex]P(X < 35) = 0.323 [/tex]

Converting to percentage

[tex]P(X < 35) = 0.323 * 100 [/tex]

[tex]P(X < 35) = 32.3 \%[/tex]

considering question b

The sample mean is [tex]\= x = 35[/tex]

Generally the standard error of the mean is mathematically represented as

[tex]\sigma_{\= x} = \frac{s}{\sqrt{n} }[/tex]

=> [tex]\sigma_{\= x} = \frac{1.3}{\sqrt{30} }[/tex]

=> [tex]\sigma_{\= x} = 0.2373 [/tex]

Generally the probability of the mean body temperature of koalas being less than 35.0°C is mathematically represented as

[tex]P(\= X < 35) = P(\frac{\= X - \mu }{\sigma_{\= x }} < \frac{35 -35.6}{0.2373 } )[/tex]

[tex]P(\= X < 35) = P(Z< -2.53 )[/tex]

From the z-table we have that

[tex]P(Z< -2.53 ) = 0.006[/tex]

So

[tex]P(\= X < 35) = 0.006 /tex]

Converting to percentage

[tex]P(\= X < 35) = 0.006 * 100 [/tex]

[tex]P(\= X < 35) = 0.6 \%[/tex]

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was [tex]P(X < 35) = 32.3 \%[/tex] but when the population was taken into account (i.e n = 30) the probability became

[tex]P(\= X < 35) = 0.6 \%[/tex]