99% of all readings are within WHAT temperature range?

Question

I am a bit stumped.

"The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean of 82.5C and standard deviation .1C. 99% of all readings are within what temperature range?"

The question states that multiple readings are taken - no sample size is given. The mean is 82.5 with standard deviation of 0.1. I am not sure whether the question is badly phrased or some other error in it.

I attempted is as a Confidence Interval, however a N would be necessary.

I would appreciate at least a hint because I feel like I am going mad.

Answer 1

Here is a different phrasing of the same problem:

Find an interval such that the probability that a reading lies within that interval is 99%.

Your thinking is too advanced and complicated. Do the simple and immediate thing.

Of course, there are many possible answers. Presumably you are asked to find the interval which is centered on the mean.

Answer 2

simply you have to calculate which is the minimum range covering 99% of the distribution. Due to the symmetry of the Gaussian you just have to waste 0.5% per tail

Can you complete the problem?

Answer 3

If $X \sim \mathsf{NORM}(\mu=82.5,\sigma=0.1),$ then you seek $L$ and $U$ for which $P(L < Y \le U) = 0.99.$ There are many solutions, but it seems easiest to cut probability $0.005$ from each tail of the distribution, as @tommik (+1) has suggested.

In R, where pnorm is a binomial CDF and pnorm is a quantile function (inverse CDF), we find $L=82.24242, U= 82.75758.$

LU=qnorm(c(.005,.995), 82.5, .1); LU
[1] 82.24242 82.75758
diff(pnorm(LU, 82.5, .1))  # check
[1] 0.99

curve(dnorm(x, 82.5, .1), 82, 83, lwd= 2, ylab="PDF", 
      xlab="x", main="Central 99% of NORM(82.1, 0.1)")
 abline(h=0, col="green2")
 abline(v=LU, lwd=2, col="red", lty="dotted")

Note: This can be done by standardizing and using printed standard normal CDF tables, but in such tables z-scores are usually given to only two places.

You might get better accuracy using the bottom row of a t table (perhaps labeled 'Inf', '$\infty$'. or 'Norm'). There you can find that $2.576$ cuts probability $0.005$ from the upper tail of a standard normal distribution.

$$P(L < X < U) = P\left(\frac{L-\mu}{\sigma} < Z =\frac{X-\mu}{\sigma} <\frac{U-\mu}{\sigma}\right)=0.99,$$ where is $Z$ is a standard normal random variable, and $\mu$ and $\sigma$ are known.

Set $(U-\mu)/\sigma = (U-82.5)/.1 = 2.576$ and solve for $U.$ Similarly for $L$ and $-2.576.$