# Mean Weight

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U= 143 lb, Stand. Dev. = 29 lb.

1a) If one woman is randomly selected, find the P that her weight is above 140 lb. I get 5517-what now???

1b) If 100 women are selected randomly, find the P that they have a mean weight greater than 140lb. (Central limit theorem)

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u= 143 lb, Stand. Dev. = 29 lb.

1a) If one woman is randomly selected, find the P that her weight is above 140 lb. I get 5517-what now???

1b) If 100 women are selected randomly, find the P that they have a mean weight greater than 140lb. (Central limit theorem)

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Please see attachment. Hope it helps.

u= 143 lb, Stand. Dev. = 29 lb.

1a) If one woman is randomly selected, find the P that her weight is above 140 lb. I get

5517-what now???

Solution. Denote the weight ...

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- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"

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