Tag Info

New answers tagged

0

Hint: $P(-z \leq Z \leq z)=2\Phi(z)-1$, with $z=\frac{x-\mu}{\sigma}$


2

First of all, see that both the distributions are symmetric - So the median would be the center value, which is $30$ Since standard deviation measures spread of the data, $B$ will be having a greater standard deviation because more data values in $B$ are away from the median compared to $A$


1

statistics tell us the Z-score (i.e., $\frac{Actual\;Lifespan - Mean\;Lifespan}{SD}$) for the elephant is -0.7 while the Z-score for the lion is 2.0. we're assuming a normal distribution and a Z score of 2.0 means that it lived longer than 97.5% of the other lions while the elephant lived only ~ 40% longer than the other elephants (look up 68-95-99.7 rule) ...


0

Cuttof value $c$ IMHO is your input by the definition. It might be 95%, 99% etc. Now you have to find a $x>0$ such that $\Phi(-x,\mu,\sigma) = 1-c$, where $x$ is such return that probability to lose more than $x$ equal $1-c$. $\mu$ is the mean, $\sigma$ is std.dev. With the bigger $\sigma$ you get the bigger $x$. $\Phi$ is cdf of normal distribution.


1

I think you're calculating things correctly. Each sampling taken separately has a standard deviation of $0$ because every item is the same price in the sample. There is no variation. But when you combine the samples into one, now you have a variation, so the standard deviation is greater than $0$.


0

We have $$ X \in N(111, 5) $$ The probability that a single match will last between 110 and 118 minutes is $$ p = P(110 \le X \le 118) = F_X(118) - F_X(110) $$ if $F_X$ is the cummulative dist. function. The probability that at least 30 (out of 64) lasts between these values is $$ 1 - \sum_{i=0}^{30} {64 \choose i}p^i(1-p)^{64-i} $$


1

Find the probability $p$ that a single match will last between 110 and 118 minutes using the normal distribution. Then use the cumulative binomial distribution to find the probability of more than 30 succeeding.


0

So this is not directly relevant to your final question, but it may help you understand a little bit as to why Standard Deviation/Variance are commonly used concepts. Variance is one of the moments of a probability distribution. I'm not sure what your math background is, so I won't go into what moments really are, but what's important to understand is that ...



Top 50 recent answers are included