Respuesta :
Answer:
a) 0.5279; b) No
Step-by-step explanation:
Since this is a normal approximation to a binomial distribution, we must use continuity correction.
The probability we are finding is P(X ≥ 121). For continuity correction, we will find
P(X > 120.5).
Our mean, μ, is 121. Our probability of success, p, is 121/201 = 0.60. Our probability of failure, q, is 1-0.60 = 0.40. Our standard deviation, σ, is
√(npq) = √(201*0.6*0.4) = √48.24 = 6.95.
The formula for a z score is
[tex]z=\frac{X-\mu}{\sigma}[/tex]. Using our information, we have
z = (120.5-121)/6.95 = -0.5/6.95 = -0.07.
Using a z table, we see that the area under the curve to the left of this is 0.4721. However, we want the area to the right; this means we subtract from 1:
1-0.4721 = 0.5279
Since this rounds to 0.53, or 53%, this does not contradict the earlier result.
As per the question [tex]53\%[/tex] of all males between age 18 and 24. To find the approximate probability we will use approximate to the binomial. Before using normal approximation to the binomial, we have to make sure the sample size is large enough.
(a) The approximate the probability that at least 121 respondents is 0.02826.
(b)
Given:
The probability is [tex]p=53\%=0.53[/tex]
Sample size is [tex]n=201[/tex].
Binomial [tex]X=121[/tex]
Calculate that mean [tex]\mu[/tex].
[tex]\mu=n\times p\\\mu=201\times 0.53\\\mu=106.53[/tex]
Calculate the standard deviation.
[tex]\sigma =\sqrt{\mu(1-p)}\\\sigma= \sqrt{106.3(1-0.53)} \\\sigma=7.06[/tex]
Write the continuity correction factor.
[tex]P(X\geq 121)=P(X\geq 120.5)\\P(X\geq 121)=P(Z\geq \frac{120.5-106.3}{7.06})\\P(X\geq 121)=P(Z\geq 2.011)\\P(X\geq 121)= 1 - P(Z < 2.011)\\P(X\geq 121)= 1 - 0.\\P(X\geq 121)=0.02\\P(X\geq 121)= 0.0286[/tex]
(b)
Yes, because the probability in part (a) is less than 0.05.
Learn more about binomial here:
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