Respuesta :
Answer:
True
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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 65, \sigma = 8, n = 64, s = \frac{8}{\sqrt{64}} = 1[/tex]
If 64 students were randomly sampled, the probability that the sample mean of the sampled students exceeds 71 minutes is approximately 0.
This probability is 1 subtracted by the pvalue of Z when X = 71. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{71 - 65}{1}[/tex]
[tex]Z = 6[/tex]
[tex]Z = 6[/tex] has a pvalue very close to 1.
1 - 1 = 0.
So approximately 0 probability that the sample mean of the sampled students exceeds 71 minutes.
The answer is true.
