Question

A random sample of n= 300 observations is selectedfrom a binomial population with p= .8. Approximateeach of the following probabilities:

1. $\mathbf{P}\left(\hat{\mathbf{p}}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{83}\right)$
   
2. $\mathbf{P}\left(\hat{\mathbf{p}}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{75}\right)$
   
3. $\mathbf{P}\left(\mathbf{0}\mathbf{.}\mathbf{79}\mathbf{<}\hat{\mathbf{p}}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{81}\right)$
   

Short answer

1. $P\left(\hat{p}<0.83\right)=0.9014$
   
2. $P\left(\hat{p}>0.75\right)=0.9850$
   
3. $P\left(0.79<\hat{p}<0.81\right)=0.3328$
   

Step-by-step solution

Given information 

We select a random sample of size n = 300 from a binomial population with a probability p = 0.8.

Calculate the probability Pp^<0.83

a.

We know that the sampling distribution is of$\hat{\mathbf{p}}$is normal according to the Central Limit Theorem with mean and standard deviation .

${\mathbf{\sigma }}_{\hat{\mathbf{p}}}\mathbf{=}\sqrt{\mathbf{p}\left(1-p\right)\mathbf{/}\mathbf{n}}$

The Z-score is given by$\frac{\hat{\mathbf{p}}\mathbf{-}{\mathbf{\mu }}_{\hat{\mathbf{p}}}}{{\mathbf{\sigma }}_{\hat{\mathbf{p}}}}$

Therefore, the probability is given by,

$\begin{array}{rcl}P\left(\hat{p}<0.83\right)& =& P\left(Z<\frac{0.83-0.8}{\sqrt{0.8\left(1-0.8\right)/300}}\right)\\ & =& P\left(Z<1.299\right)\\ & \approx & 0.9014\\ & & \end{array}$

We get the probability approximately 90% from the z-score table.

Thus, the required probability is 0.9014.

Calculate the probability Pp^>0.75

b.

The probability is given by,

$\begin{array}{rcl}P\left(\hat{p}>0.75\right)& =& P\left(Z>\frac{0.75-0.8}{\sqrt{0.8\left(1-0.8\right)/300}}\right)\\ & =& P\left(Z>-2.17\right)\\ & \approx & 0.9850\\ & & \end{array}$

We get the probability approximately 98% from the z-score table.

Thus, the required probability is 0.9850.

Calculate the probability P0.79<p^<0.81

c.

The probability is given by,

$\begin{array}{rcl}P\left(0.79<\hat{p}<0.81\right)& =& P\left(\frac{0.79-0.8}{\sqrt{0.8\left(1-0.8\right)/300}}<Z<\frac{0.81-0.8}{\sqrt{0.8\left(1-0.8\right)/300}}\right)\\ & =& P\left(-0.43<Z<0.43\right)\\ & =& P\left(Z<0.43\right)-P\left(Z<-0.43\right)\\ & =& 0.6664-0.3336\\ & =& 0.3328\\ & & \end{array}$

Thus, we get the probability approximately 33%from the z-score table.

Thus, the required probability is 0.8828.