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.
