Answer:
a) 0 .0355
b) 0 .0345
c) The probabilities differ by 0.0010
Step-by-step explanation:
We are given the following information:
a) P(Success) = 0.3
n = 58
x = 12
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 58 and x = 12
We have to evaluate:
[tex]P(x = 12) = \binom{58}{12}(0.3)^{12}(1-0.3)^{46} = 0.0355[/tex]
b) Using the normal distribution
n = 12
p = 0.3
q = 1 - p = 1 - 0.3 = 0.7
b) [tex]\text{Mean}=np = 12\times 0.3 = 3.6\\ \text{Standard Deviation} = \sqrt{npq} = \sqrt{12\times 0.3\times 0.7} = 1.58[/tex]
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(x = 12)
[tex]P( x \leq 12)-P(x < 12[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x = 12) = 0.0345[/tex]
c) Difference
0.0355 - 0.0345 = 0.0010
Thus, the probabilities differ by 0.0010