[tex]n_A p_A = 6\\n_A (1 - p_A) = 24\\n_B p_B = 5.4\\n_B (1 - p_B) = 24.6[/tex]
The sampling distribution of the sample proportions is approximately normal.
Check the Large Counts Condition:
The large counts condition is satisfied if both [tex]n_A p_A[/tex] and [tex]n_A (1 - p_A)[/tex] are at least 10. In this case, we are given that [tex]n_A = 30[/tex] and [tex]p_A = 0.20[/tex]. Substituting these values into the condition, we get:
[tex]n_A p_A = (30) (0.20) = 6\\n_A (1 - p_A) = (30) (0.80) = 24[/tex]
Since both values are greater than 10, the large counts condition is satisfied.
Calculate the values:
We are asked to find the values of:
[tex]n_A p_A[/tex] (which we already calculated as 6)
[tex]n_A (1 - p_A)[/tex] (which we already calculated as 24)
[tex]n_B p_B[/tex] (Similarly, this is [tex]n_B p_B = (30) (0.18) = 5.4[/tex])
[tex]n_B (1 - p_B)[/tex] (similarly, this is [tex]n_B (1 - p_B) = (30) (0.82) = 24.6[/tex])
Determine the shape of the sampling distribution:
Since large counts condition is satisfied, the sampling distribution of the sample proportions will be approximately normal.