This question is incomplete, the complete question is;
In two shipments of fruit, A and B, the proportions of fruit that are apples are [tex]P^"_{A}[/tex] and [tex]P^"_{B}[/tex] , respectively. Suppose that independent random samples of 50 fruit from A and 100 fruit from B. Let [tex]P^"_{A}[/tex] Â be the sample proportion of apples from shipment A and [tex]P^"_{B}[/tex] be the sample proportion of apples from shipment B. What is the mean of the sampling distribution of [tex]P^"_{A}[/tex] - [tex]P^"_{B}[/tex] Â ?
Answer:
the mean of the sampling distribution is [tex]P^"_{A}[/tex] - [tex]P^"_{B}[/tex] Â
Step-by-step explanation:
Given the data in the question;
let [tex]X_{A}[/tex] be number of apples from the shipment A
and [tex]X_{B}[/tex] be number of apples from shipment B
[tex]P^"_{A}[/tex]  =  [tex]X_{A}[/tex] /  [tex]n_{A}[/tex]  =  [tex]X_{A}[/tex] / 50               {  x ¬ Bin(np )
{ [tex]X_{A}[/tex] = 50[tex]P^"_{A}[/tex] Â )
[tex]P^"_{B}[/tex] Â = Â [tex]X_{B}[/tex] / Â [tex]n_{B}[/tex] Â = Â [tex]X_{B}[/tex] / 100
( [tex]X_{B}[/tex] = 100[tex]P^"_{B}[/tex] )
E( [tex]P^"_{A}[/tex] - [tex]P^"_{B}[/tex] Â ) = E( [tex]X_{A}[/tex] / Â [tex]n_{A}[/tex] Â - [tex]X_{B}[/tex] / Â [tex]n_{B}[/tex] )
= E( [tex]X_{A}[/tex] / 50 Â - Â [tex]X_{B}[/tex] / 100 )
= E([tex]X_{A}[/tex]) / 50 Â - Â E([tex]X_{B}[/tex]) / 100
= 50[tex]P^"_{A}[/tex] / 50 Â - Â 100[tex]P^"_{B}[/tex] Â / 100
= Â [tex]P^"_{A}[/tex] - [tex]P^"_{B}[/tex] Â Â Â Â Â Â { E(X) = np ]
Therefore, the mean of the sampling distribution is [tex]P^"_{A}[/tex] - [tex]P^"_{B}[/tex] Â
