Respuesta :
Answer:
a) 10.38% probability that the sample mean will be more than 59 pounds.
b) 67.72% probability that the sample mean will be more than 56 pounds.
c) 22.10% probability that the sample mean will be between 56 and 57 pounds.
d) 1.46% probability that the sample mean will be less than 53 pounds.
e) 0% probability that the sample mean will be less than 49 pounds.
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 question, we have that:
[tex]\mu = 56.8, \sigma = 12.2, n = 49, s = \frac{12.2}{\sqrt{49}} = 1.74285[/tex]
a. More than 59 pounds
This is 1 subtracted by the pvalue of Z when X = 59. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{59 - 56.8}{1.74285}[/tex]
[tex]Z = 1.26[/tex]
[tex]Z = 1.26[/tex] has a pvalue of 0.8962.
1 - 0.8962 = 0.1038
10.38% probability that the sample mean will be more than 59 pounds.
b. More than 56 pounds
This is 1 subtracted by the pvalue of Z when X = 56. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{56 - 56.8}{1.74285}[/tex]
[tex]Z = -0.46[/tex]
[tex]Z = -0.46[/tex] has a pvalue of 0.3228.
1 - 0.3228 = 0.6772
67.72% probability that the sample mean will be more than 56 pounds.
c. Between 56 and 57 pounds
This is the pvalue of Z when X = 57 subtracted by the pvalue of Z when X = 56. So
X = 57
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{57 - 56.8}{1.74285}[/tex]
[tex]Z = 0.11[/tex]
[tex]Z = 0.11[/tex] has a pvalue of 0.5438
X = 56
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{56 - 56.8}{1.74285}[/tex]
[tex]Z = -0.46[/tex]
[tex]Z = -0.46[/tex] has a pvalue of 0.3228.
0.5438 - 0.3228 = 0.2210
22.10% probability that the sample mean will be between 56 and 57 pounds.
d. Less than 53 pounds
This is the pvalue of Z when X = 53.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{53 - 56.8}{1.74285}[/tex]
[tex]Z = -2.18[/tex]
[tex]Z = -2.18[/tex] has a pvalue of 0.0146
1.46% probability that the sample mean will be less than 53 pounds.
e. Less than 49 pounds
This is the pvalue of Z when X = 49.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{49 - 56.8}{1.74285}[/tex]
[tex]Z = -4.48[/tex]
[tex]Z = -4.48[/tex] has a pvalue of 0.
0% probability that the sample mean will be less than 49 pounds.
