Answer:
a
normal approximation not appropriate in this instance because
[tex]np = 3300 * 0.9998 = 3299.3 > 5[/tex]
and
[tex]nq = 3300 * 0.0002 = 0.66 < 5[/tex]
b
The probability is [tex]P(X = 3300) = 0.52[/tex]
c
The probability is [tex]P(X = 3287) = 0[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 3300[/tex]
The chance of liking blueberry pie is [tex]p = 0.9998[/tex]
The chance of not liking blueberry pie is [tex]q = 0.0002[/tex]
For normal approximation is possible if
[tex]np > 5[/tex] , [tex]nq > 5[/tex]
Now let test
[tex]np = 3300 * 0.9998 = 3299.3 > 5[/tex]
and
[tex]nq = 3300 * 0.0002 = 0.66 < 5[/tex]
The probability that all 3300 people will say they like blueberry pie is mathematically represented as
[tex]P(X = 3300) = p^{3300}[/tex]
[tex]P(X = 3300) = (0.9998)^{3300}[/tex]
[tex]P(X = 3300) = 0.52[/tex]
The probability that 3297 of the people will say that they like blueberry pie is mathematically represented as
[tex]P(X = 3287) = \frac{(np)^{3297! } * e^{-np}}{3297}[/tex]
[tex]P(X = 3287) = \frac{0}{\infty}[/tex]
[tex]P(X = 3287) = 0[/tex]
