Question

Please show the steps and if you could solve it on paper and then upload a picture that would be awesome, fractions and formulas can get a bit distorted when typed.

.....................

a.The student health center at a university treats an average of seven cases of mononucleosis per day during the week of final examinations. Find the probability that on a given day during the finals week exactly four cases of mononucleosis will be treated at this health center.

b. The amounts of electricity bills for all households in a city have skewed probability distribution with a mean of $140 and a Standard deviation of $30. Find the probability that the mean amount of electric bills for a random sample of 75 households selected from this city will be within $6 of the population mean.  

Answer 1

Solution:

    a) we are given that : the student health center at a university treats an average of seven cases of mononucleosis per day during the week of final examinations.

   We have to find the probability that : on a given day during the finals week exactly four cases of mononucleosis will be treated at this health center.

x = number of cases of mononucleosis per day during the week of final examinations follows Poisson distribution with parameter $\lambda =7$

thus pmf of x is :

$P(X=x)=\frac{e^{-\lambda}\times \lambda ^{x}}{x!}$              x = 0,1,2,3,...............

We have to find :
P( X = 4) = .......?

thus

$P(X=4)=\frac{e^{-7}\times 7 ^{4}}{4!}$

Using scientific calculator find e^-7  

e^-7   =0.000912

or use excel

=EXP(-7)

Thus we get :

$P(X=4)=\frac{0.000912 \times 2401}{4\times 3\times 2\times 1}$

$P(X=4)=\frac{2.189429 }{24}$

$P(X=4)=0.0912$

b) The amounts of electricity bills for all households in a city have skewed probability distribution with a mean of $140 and a Standard deviation of $30.

Thus Mean = $\mu = 140$

Standard deviation = $\sigma = 30$

A random sample of 75 households selected from this city

Thus sample size = n = 75

and we have to find the probability that : the mean amount of electric bills for a random sample of 75 households selected from this city will be within $6 of the population mean.  

That is :

$P(\mu-6\leq \bar{x}\leq \mu +6)=........?$

Since sample size n = 75 > 30, we can assume it is large sample and hence using central limit theorem , we assume Approximate Norma distribution and we will find above probability.

$P(\mu-6-\mu\leq( \bar{x} -\mu )\leq \mu +6-\mu)=........?$

divide by $\sigma /\sqrt{n}$

$P(\frac{-6}{\sigma /\sqrt{n}}\leq\frac{( \bar{x} -\mu )}{\sigma /\sqrt{n}}\leq \frac{6}{\sigma /\sqrt{n}})=........?$

$P(\frac{-6}{30 /\sqrt{75}}\leq Z \leq \frac{6}{30 /\sqrt{75}})=........?$

$P(\frac{-6}{30 /8.66025 }\leq Z \leq \frac{6}{30 /8.66025 })=........?$

$P(\frac{-6}{3.46410 }\leq Z \leq \frac{6}{3.46410 })=........?$

$P( - 1.73 \leq Z \leq 1.73 )=........?$

$P( - 1.73 \leq Z \leq 1.73 )=P( Z \leq 1.73) - P( Z \leq - 1.73)$

Look in z table for z = 1.7 and 0.03 as well as for z = -1.7 and 0.03 and find area.

P( Z < -1.73 ) = 0.0418

P( Z < 1.73 )= 0.9582

Thus we get :

$P( - 1.73 \leq Z \leq 1.73 )=0.9582 - 0.0418$

$P( - 1.73 \leq Z \leq 1.73 )=0.9164$

Thus the probability that the mean amount of electric bills for a random sample of 75 households selected from this city will be within $6 of the population mean is 0.9164.