Suppose that 30% of female students and 25% of male students at a large high school are enrolled in an ap ® class. independent random samples of 20 females and 20 males are selected and asked if they are enrolled in an ap ® class. let ˆ p f represent the sample proportion of females enrolled in an ap ® class and ˆ p f represent the sample proportion of males enrolled in an ap ® class. what is the mean of the sampling distribution of ˆ p f − ˆ p m ?

The mean of the sampling distribution of (ˆpF − ˆpM) = 0.05

Step-by-step explanation:

According to the central limit theorem, a sample extracted from a set of data is said to approximate a normal distribution with its sampling distribution having the same mean/proportion as the population mean/proportion.

Hence,

The sample proportion of females enrolled in an AP ® class = The population proportion of females enrolled in an AP ® class

ˆpF = 30% = 0.30

And the sample proportion of males enrolled in an AP ® class = The population proportion of males enrolled in an AP ® class

ˆpM = 25% = 0.25

So, the mean of the sampling distribution of (ˆpF − ˆpM) will simply be 30% - 25% = 5% = 0.05

Hope this Helps!!!

okay so 15.15 *()5: +5/6

step-by-step explanation:

$53.50

step-by-step explanation:

you have to subtract the amount of money chuck earned ($72.50) by the cost of the gas for the mower ($11.25) and the new string for the trimmer he bought ($7.75).

72.50-11.25=61.25

and then subtract $61.25 by the second thing he bought (the new string for the trimmer)

61.25-7.75=53.5

so in total it's $53.50 (53 dollars and 50 cents)