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Sagot :
Answer:
a)
0.3632 = 36.32% approximate probability that 15 or fewer turn right.
0.369 = 36.9% exact probability that 15 or fewer turn right.
b)
0.4801 = 48.01% approximate probability that at least two-thirds of those in the sample turn.
0.4868 = 48.68% exact probability that at least two-thirds of those in the sample turn.
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x sucesses 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 zscore 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 pvalue, 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].
Vehicles entering an intersection from the east are equally likely to turn left, turn right, or proceed straight ahead.
This means that [tex]p = \frac{1}{3}[/tex]
50 vehicles
This means that [tex]n = 50[/tex]
Mean and standard deviation:
[tex]\mu = E(X) = np = 50\frac{1}{3} = 16.67[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50\frac{1}{3}\frac{2}{3}} = 3.33[/tex]
a. 15 or fewer turn right.
Using continuity correction, this is [tex]P(X \leq 15 + 0.5) = P(X \leq 15.5)[/tex], which is the pvalue of Z when X = 15.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{15.5 - 16.67}{3.33}[/tex]
[tex]Z = -0.35[/tex]
[tex]Z = -0.35[/tex] has a pvalue of 0.3632
0.3632= 36.32% approximate probability that 15 or fewer turn right.
Using a binomial probability calculator, to find the exact probability, we get a 0.369 = 36.9% exact probability that 15 or fewer turn right.
b. at least two-thirds of those in the sample turn.
Turn either left or right, so:
[tex]p = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}[/tex]
The standard deviation remains the same, while the mean will be:
[tex]\mu = E(X) = np = 50\frac{2}{3} = 33.33[/tex]
Two thirds of the sample is 33.33, so at least 34 turning, which, using continuity correction, is [tex]P(X \geq 34 - 0.5) = P(X \geq 33.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 33.5.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{33.5 - 33.33}{3.33}[/tex]
[tex]Z = 0.05[/tex]
[tex]Z = 0.05[/tex] has a pvalue of 0.5199
1 - 0.5199 = 0.4801
0.4801 = 48.01% approximate probability that at least two-thirds of those in the sample turn.
Using a binomial probability calculator, we find a 0.4868 = 48.68% exact probability that at least two-thirds of those in the sample turn.
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