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.

The assignment is given:

In a heat regulated room, we have two temperature limits $T_{\text{min}} = 16$ and $T_{\text{max}} = 24$. If the temperature is between $T_{\text{min}}$ and $T_{\text{max}}$, the temperature regulation system is off.

Let's assume that if there was no heat regulation in the room, the temperature would be normal distributed, with then mean value $\mu = 21$ (Celsius) and the standard deviation $\sigma = 2.7$.

1) How much percentage of time, is the heat regulation system off?

Let's assume that the room air circulation is improved, so that the temperature regulation system now if off $92\%$ of time, with the same values, $\mu$, $T_{\text{min}}$ and $T_{\text{min}}$ as in (1).

2) What is the standard deviation ($\sigma$) now?

Using R, I have found that the heat regulation is off about $16.2\%$ of the time, using the following command.

1 - pnorm(24,21,2.7) - pnorm(16,21,2.7)

This is correct, I know.

However, in the latter problem, I quickly ran into problems. I asked my professor, and he could not figure it out either.

Anyone here, willing to help me out?

share|improve this question

3 Answers 3

HINT: Heat regulation amounts to truncation of room temperatures to the preset temperature window, i.e. $16 \le T \le 24$. The distribution of the temperature is now truncated normal, i.e. it has density $$ f_T(t) = \frac{1}{\sqrt{2 \pi} \sigma} \exp\left(-\frac{(t-\mu)^2}{2 \sigma^2}\right) \frac{1}{ \Phi\left(\frac{T_\mathrm{max}-\mu}{\sigma}\right) - \Phi\left(\frac{T_\mathrm{min}-\mu}{\sigma}\right) } \mathbf{1}\left( T_\mathrm{min} \le t \le T_\mathrm{max} \right) $$

You could work the standard deviation out numerically, using this density.

But it is intuitively clear that it would become smaller than the original $\sigma$.

share|improve this answer

It's much easier for $\mu=20$:

Let $T$ be the temperature of the room. You are told that $$ P[16< T <24]=.92. $$

Pass to the standard normal variable:

$$ P[{-4\over \sigma}< Z <{4\over \sigma} ] =.92. $$

This is equivalent to $$ 1-2P[Z> {4\over \sigma}] =.92 $$ or $$ P[Z> {4\over \sigma}] =0.04. $$

Doing a ``reverse lookup'' in a cdf table for the normal distribution gives $$ {4\over \sigma} =1.751. $$ So $\sigma=4/1.751=2.2844 $.

(all numerical quantities approximate.)

With $\mu=21$, things become difficult, since we can't take advantage of symmetry as in the second step above. You'd have to ''guess and check'', using the reverse lookup (for $P[{-5\over \sigma}< Z < {3\over\sigma}]=0.92$) to approximate $\sigma$.

share|improve this answer
This ignores Sasha's observation that the temperature has a truncated normal distribution. You'd have to use his suggestion to get the exact answer. I imagine this would be tedious, though... –  David Mitra Nov 8 '11 at 17:19

You may have noticed that for standard deviation $2.7$, the left tail contributes very little to the probability. With better temperature regulation, the standard deviation will be smaller, so the left tail will contribute even less, indeed negligibly little.

Note that from tables, or something fancier, you can find that there is $8\%$ in the right tail $Z \ge z$ of the standard normal if $z\approx 1.41$. More precisely, this is the first place in my primitive table where the area in the right tail dips below $8\%$. In fact, the probability that $Z\ge 1.41$ is about $0.0793$. Thus $24$ degrees is about $1.41$ standard deviation units from the mean. If the standard deviation is $\sigma$, we get $$1.41\sigma\approx 3. \qquad(\ast)$$ With this $\sigma$, the left tail is $2.82$ standard deviation units away from the mean, so contributes roughly $0.0025$ to our desired probability of $8\%$. Add. We find that for the $\sigma$ obtained from $(\ast)$, the probability of being off is about $0.0793+0.0025$, which is $0.0818$, a little too big.

Back to the table! We want to find the $z$ that gives right tail area of about $0.08-0.0025=0.0775$. This is at $z\approx 1.42$, not a big change from $1.41$. So solving $$1.42\sigma=3$$ will give us a near perfect $8\%$.

Comment: Probably your software can solve automatically for $\sigma$. The equation is $$P(Z \ge 3/\sigma)+P(Z \le -6/\sigma)=0.08.$$

share|improve this answer

Your Answer


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.