Respuesta :
Answer:
In 93.12% of the samples the sample mean height would be between 135 and 155 cm
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples 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.
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 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 = 145, \sigma = 22, n = 16, s = \frac{22}{\sqrt{16}} = 5.5[/tex]
In what percentage of the samples would the sample mean height be between 135 and 155 cm?
This is the pvalue of Z when X = 155 subtracted by the pvalue of Z when X = 135. So
X = 155
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{155 - 145}{5.5}[/tex]
[tex]Z = 1.82[/tex]
[tex]Z = 1.82[/tex] has a pvalue of 0.9656
X = 135
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{135 - 145}{5.5}[/tex]
[tex]Z = -1.82[/tex]
[tex]Z = -1.82[/tex] has a pvalue of 0.0344
0.9656 - 0.0344 = 0.9312
In 93.12% of the samples the sample mean height would be between 135 and 155 cm
