Respuesta :
Answer:
a) 0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51
b) 0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51
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 = 50, \sigma = 1.8[/tex]
(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 17 pins is at least 51?
Here [tex]n = 17, s = \frac{1.8}{\sqrt{17}} = 0.4366[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 51. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{51 - 50}{0.4366}[/tex]
[tex]Z = 2.29[/tex]
[tex]Z = 2.29[/tex] has a pvalue of 0.9890
1 - 0.989 = 0.011
0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51
(b) What is the (approximate) probability that the sample mean hardness for a random sample of 45 pins is at least 51?
Here [tex]n = 17, s = \frac{1.8}{\sqrt{45}} = 0.2683[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{51 - 50}{0.0.2683}[/tex]
[tex]Z = 3.73[/tex]
[tex]Z = 3.73[/tex] has a pvalue of 0.9999
1 - 0.9999 = 0.0001
0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51
