Respuesta :
Answer:
P(r ≥ 15) = 0.9943.
Step-by-step explanation:
We use the normal approximation to the binomial to solve this question.
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
625 trials and a probability of a 4.4% success on a single trial.
This means that [tex]n = 625, p = 0.044[/tex]
Mean and standard deviation:
[tex]mu = E(X) = np = 625*0.044 = 27.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{625*0.044*0.956} = 5.13[/tex]
P(r ≥ 15)
Using continuity correction, this is [tex]P(r \geq 15 - 0.5) = P(r \geq 14.5)[/tex], which is 1 subtracted by the p-value of Z when X = 14.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{14.5 - 27.5}{5.13}[/tex]
[tex]Z = -2.53[/tex]
[tex]Z = -2.53[/tex] has a p-value of 0.0057
1 - 0.0057 = 0.9943
So
P(r ≥ 15) = 0.9943.
