help with stats! A variable X has a normal distribution with mean 100. It is known that about 47.5% of the values of X fall between 85 and 100. What is the approximate value of the standard deviation ?
see attachment:
@kropot72 @agent0smith
A recycling plant compresses aluminum cans into bales. The weights of the bales are known to follow a normal distribution with standard deviation eight pounds. In a ran- dom sample of 64 bales, what is the probability that the sample mean diers from the population mean by no more than one pound? see attachment
A variable X has a normal distribution with mean 100\[ \Pr(X\leq 100) = 0.5 \] It is known that about 47.5% of the values of X fall between 85 and 100.\[ \Pr(X> 85, X \leq 100) = 0.457 \]
When we simplify we get: \[ 1 - \bigg (\Pr(X\leq 85) + \Pr(X>100) \bigg) = 0.475 \\ 1 - \Pr(X \leq 85) - 0.5 = 0.475 \\ \Pr(X\leq 85) = 0.5 - 0.475 = 0.025 \]
\[ \large \Phi \left(\frac{85-100}{\sigma}\right) = 0.025\\ \large \Phi \left(\frac{-15}{\sigma}\right) = 0.025 \]Or: \[ \sigma = \frac{-15}{\Phi^{-1}(0.025)} \]
Recycling plant question: According to the Central Limit Theorem the distribution of the sample means is approximately normal, with standard deviation \[\frac{\sigma}{\sqrt{n}}=\frac{8}{\sqrt{64}}=1\ pound\] In this case the required probability is found by first finding the cumulative probability for z = 1 using a standard normal distribution table. Then subtract from that value the cumulative probability for z = -1.
thanks@kropot72 and @wio
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