Respuesta :
Answer:
a) The probability is 0.1170
b) The probability is 0.7405
c) D. The normal distribution can be used because the original population has a normal distribution.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are 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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 195, \sigma = 7.8[/tex]
a. Find the probability that an individual distance is greater than 204.30 cm.
This is 1 subtracted by the pvalue of Z when X = 204.30. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{204.30 - 195}{7.8}[/tex]
[tex]Z = 1.19[/tex]
[tex]Z = 1.19[/tex] has a pvalue of 0.8830
1 - 0.8830 = 0.1170
11.70% probability that an individual distance is greater than 204.30 cm.
b. Find the probability that the mean for 15 randomly selected distances is greater than 193.70 cm.
Now we have [tex]n = 15, s = \frac{7.8}{\sqrt{15}} = 2.014[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 193.70.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{193.70 - 195}{2.014}[/tex]
[tex]Z = -0.645[/tex]
[tex]Z = -0.645[/tex] has a pvalue of 0.2595
1 - 0.2595 = 0.7405
The probability is 0.7405
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
Because the population distribution is normal, so the correct answer is:
D. The normal distribution can be used because the original population has a normal distribution.
