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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.

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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 $$\Pr(X \gt 370)=\Pr\left(Z\gt \frac{370-300}{35}\right).$$

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