Respuesta :
Answer:
0.6064 = 60.64% probability that fewer than 29 are left-handed.
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].
16% of the population of the U.S. is left-handed.
This means that [tex]p = 0.16[/tex]
Sample of 170 people
This means that [tex]n = 170[/tex]
Mean and standard deviation:
[tex]\mu = E(X) = np = 170*0.16 = 27.2[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{170*0.16*0.84} = 4.78[/tex]
Probability that fewer than 29 are left-handed.
Using continuity correction, this is P(X < 29 - 0.5) = P(X < 28.5), which is the pvalue of Z when X = 28.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{28.5 - 27.2}{4.78}[/tex]
[tex]Z = 0.27[/tex]
[tex]Z = 0.27[/tex] has a pvalue of 0.6064
0.6064 = 60.64% probability that fewer than 29 are left-handed.
