Question 754925: . (a) With n = 14 and p = 0.3, find the binomial probability P(9) by using a binomial probability table.
(b) If np ≥5 and nq ≥5, also estimate the indicated probability by using the normal distribution as an approximation to the binomial; if np < 5 or nq < 5, then state that the normal approximation cannot be used.
(a) Find the probability by using a binomial probability table.
P(9) = [ ] (Round to four decimal places as needed).
(b) Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. P(9) = [ ] (Round to four decimal places as needed)
B. The normal distribution cannot be used.
Also,
If the random variable z is a Standard Normal Score, what is P(-2.00 ≤ z ≤ +2.00)? How did you find this probability?
Find the z score for the standard normal distribution where: P(z<+a) = 0.9625
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (a) With n = 14 and p = 0.3, find the binomial probability P(9) by using a binomial probability table.
P(x = 9) = 14C9*(0.3)^9*(0.7)^5 = binompdf(14,0.3,9) = 0.0066
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(b) If np ≥5 and nq ≥5, also estimate the indicated probability by using the normal distribution as an approximation to the binomial; if np < 5 or nq < 5, then state that the normal approximation cannot be used.
mean = np = 14*0.3 = 4.2 ; std = sqrt[npq] = sqrt(4.2*0.7) = 1.7146
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P(x = 9) = P(8.5< x <9.5)
Find the z-values:
z(8.5) = (8.5-4.2)/1.7146 = 2.5079
z(9.5) = (9.5-4.2)/1.7146 = 3.0911
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= P(2.5079< z <3.0911) = 0.0051
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Also,
If the random variable z is a Standard Normal Score, what is P(-2.00 ≤ z ≤ +2.00)? How did you find this probability?
Ans: Using a TI-84 I get: normalcdf(-2,2) = 0.9545
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Find the z score for the standard normal distribution where: P(z<+a) = 0.9625
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Find the z-value with a left tail of 0.9625
a = invNorm(0.9625) = 1.0875
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cheers,
Stan H.
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