We appreciate your visit to Assume that weights of men are normally distributed with a mean of 170 lb and a standard deviation of 30 lb approximate values based on. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Answer:
a) [tex]P(X<185)=P(Z<0.5)=0.691[/tex]
b) [tex]P(\bar X <185)=P(Z<3)=0.99865[/tex]
Step-by-step explanation:
1) Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
2) Part a
Let X the random variable that represent the weights of men of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(170,30)[/tex]
Where [tex]\mu=170[/tex] and [tex]\sigma=30[/tex]
We are interested on this probability
[tex]P(X<185)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<185)=P(\frac{X-\mu}{\sigma}<\frac{185-\mu}{\sigma})=P(Z<\frac{185-170}{30})=P(Z<0.5)[/tex]
And we can find this probability on this way:
[tex]P(Z<0.5)=0.691[/tex]
3) Part b
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex]P(\bar X <185)=P(Z<\frac{185-170}{\frac{30}{\sqrt{36}}}=3)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<3)=0.99865[/tex]
Thanks for taking the time to read Assume that weights of men are normally distributed with a mean of 170 lb and a standard deviation of 30 lb approximate values based on. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
Final answer:
The question involves calculating the percentage of individual men with weights under 185 lb and the percentage of the sample means under 185 lb using the concepts of Z-score and the Central Limit Theorem.
Explanation:
The question is about finding the percentage of men with weights less than 185 lb and the percentage of the sample means less than 185 lb. We first address the individual weights which are normally distributed with a mean (μ) of 170 lb and a standard deviation (σ) of 30 lb. To find the percentage of individual men weighing less than 185 lb, we calculate the Z-score using the formula Z = (X - μ) / σ, where X is 185 lb. The Z-score obtained is then used to find the corresponding percentage from the standard normal distribution table.
For the second part of the question, since samples of 36 men are selected, we use the Central Limit Theorem. The sample mean also follows a normal distribution with the same mean (μ) of 170 lb, but a reduced standard deviation (σ/√n), where n is the sample size (36 in this case). We then recalculate the Z-score for the sample mean being less than 185 lb and find the corresponding percentage from the standard normal distribution table.
