Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

1.in a normal distribution data the standard deviation is greater than the quartile deviation and the mean deviation ??

  1. in a normal distribution 31% of the items are under 45 and 8% are over 64 find the mean and standard deviation ?

  2. the life time of a certain kind of battery has a mean of 7 300 hours and a standard deviation 35 hrs .assuming the distribution of life times which are measured which have life time of more than 370 hrs.

share|improve this question
2  
Are these homework questions? If so, you should indicate that by adding the (homework) tag. Additionally, these all seem very straightforward---what's giving you problems? What have you tried? What terms don't you understand? –  Rick Decker Jul 25 '12 at 13:48
    
well nope these are not homework questions but something which im expecting in maths exams ,if i knew them i wouldnt have posted man. –  user36495 Jul 25 '12 at 14:53
    
This is an answer to @Rick's first question. What about his three other questions? –  Did Jul 25 '12 at 15:05

1 Answer 1

"Standard deviation," "quartile deviation," and "mean deviation" are all measures of dispersion, of how spread out a distribution is.

Standard deviation is the most familiar. Quartile deviation is half the difference between the third quartile and the first quartile. In the standard normal $Z$, the third quartile, that is, the place $k$ such that $\Pr(Z \le k)=0.75$, is approximately $0.675$. (I got this by looking at a table of the standard normal.) The first quartile is at $-0.675$, and so the difference divided by $2$ is approximately $0.675$. In the general normal with standard deviation $\sigma$, the quartile deviation is approximately $0.675\sigma$. In particular, this is substantially less than the standard deviation $\sigma$.

The mean deviation is the expectation of $|X-\mu|$. For the standard normal we need therefore to calculate $\int_{-\infty}^\infty |z| f(z)\,dz$, where $f(z)$ is the density function of the standard normal. This integral turns out to be easy to calculate. It is $\sqrt{2/\pi}$. For the normal with standard deviation $\sigma$, the mean deviation is $\sqrt{2/\pi}\sigma$. The calculator shows this is about $0.798\sigma$.


For Problem $1$: Let our random variable be $X$, let $\mu$ be the mean and let $\sigma$ be the standard deviation. We are told that with $X \lt 45$ with probability $0.31$.

The number $z$ such that $\Pr(Z \le z)=0.31$ is approximately $-0.495$. I have a feeling you are expected to use the approximation $0.5$. The place $z$ such that $\Pr(Z \gt z)=0.08$ is approximately $1.405$. Probably you are expected to use $1.4$.

So $45$ is about $0.5$ standard deviation units below the mean, and $64$ is about $1.4$ standard deviation units above the mean. In equations, $$45=\mu-0.5\sigma\quad\text{and}\quad 64=\mu+1.4\sigma.$$ Now we can solve these linear equations for $\mu$ and $\sigma$. The answers turn out to be exceptionally simple, but you should remember the are approximations.

For Problem $2$: The problem has typos, but seems to ask for the probability that a normal with mean $300$, standard deviation $35$, is greater than $370$. But $370$ is $2$ standard deviation units above the mean $300$.

More formally, if we let our random variable be $X$, then $$\Pr(X \gt 370)=\Pr\left(Z\gt \frac{370-300}{35}\right).$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.