Respuesta :
Answer:
a) 89.05% probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200
b) No, because one of the requirements of the central limit theorem is a sample size of at least 30.
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 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 = 10000, \sigma = 500[/tex]
(a) What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200?
Here we have [tex]n = 40, s = \frac{500}{\sqrt{40}} = 79.06[/tex]
This probability is the pvalue of Z when X = 10200 subtracted by the pvalue of Z when X = 9900. So
X = 10200
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{10200 - 10000}{79.06}[/tex]
[tex]Z = 2.53[/tex]
[tex]Z = 2.53[/tex] has a pvalue of 0.9943.
X = 9900
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{9900 - 10000}{79.06}[/tex]
[tex]Z = -1.26[/tex]
[tex]Z = -1.26[/tex] has a pvalue of 0.1038.
0.9943 - 0.1038 = 0.8905
89.05% probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200
(b) If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information?
No, because one of the requirements of the central limit theorem is a sample size of at least 30.
Following are the solution to the given expression:
Given:
[tex]\text{mean} \ (\mu) = 10,000\\\\\text{standard deviation} \ ( \sigma) = 500[/tex]
Solution:
For point a:
[tex]\to n = 40 \\\\\to \mu_{\bar{x}} = \mu = 10,000\\\\\to \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{500}{\sqrt{40}} = 79.06\\\\\to P( 9,900 < \bar{x}< 10,200 )\\\\[/tex]
[tex]\to P[ \frac{( 9900 - 10000 )}{79.06} < \frac{(\bar{x} - \mu_{\bar{x}} )}{\sigma_{\bar{x}}} < \frac{P( 10200 - 10000 )}{79.06} )] \\\\[/tex]
[tex]\to P( -1.26 < Z < 2.53 )\\\\\to P(Z < 2.53 ) - P(Z < -1.26 )\\\\[/tex]
Using z table
[tex]\to 0.9943 - 0.1038 \\\\\to 0.8905[/tex]
For point b:
[tex]\to n = 15\\\\\to \mu_{\bar{x}} = \mu = 10000\\\\ \to \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{500}{\sqrt{15}} = 129.10\\\\\to P( 9,900 < \bar{x} < 10,200 )[/tex]
[tex]\to P[\frac{( 9900 - 10000)}{129.10} < ( \frac{\bar{x} -\mu_{\bar{x}})}{\sigma_{\bar{x}}} < \frac{( 10200 - 10000 )}{129.10} )]\\\\\to P( -0.77 < Z < 1.55 )\\\\ \to P(Z < 1.55 ) - P(Z < -0.77)[/tex]
Using z table
[tex]\to 0.9394 - 0.2206\\\\ \to 0.7188[/tex]
Learn more about the probability:
brainly.com/question/13708313
