Question

The heights of seedlings are normally distributed. Given that 10% of the seedlings are taller than 15 cm and 5% are shorter than 4 cm, find the mean and standard deviation of the heights.​

Answer 1

There are many possible ranges, including (−∞,15.051464)

and (15.051464,18.6844845)

as well as 31.5±0.6270678

(centered on the mean). Observe, in R:

> pnorm(15.051464, 31.5, 10) - pnorm(-Inf, 31.5, 10)

[1] 0.05

> pnorm(18.6844845, 31.5, 10) - pnorm(15.051464, 31.5, 10)

[1] 0.05

> pnorm(32.1270678, 31.5, 10) - pnorm(30.8729322, 31.5, 10)

[1] 0.05

In general, if you are given a small enough but otherwise arbitrary number a∈R

you can find a unique b

such that b>a

and

∫baϕ(x,31.5,10)dx=0.05,

where ϕ

is the normal density. For example, if a=10

, then b≈16.42

. Again in R:

> qnorm(pnorm(10, 31.5, 10) + 0.05, 31.5, 10)

[1] 16.42003

So your question doesn’t determine the answer uniquely: I’m free to choose the upper or lower bound of the interval or any particular point relative to those (e.g. the midpoint) arbitrarily.

If you had asked for a 95% interval, it would have been a reasonable assumption (but an assumption nonetheless) that you want an interval centered on the mode, median, or mean. But for a 5% interval, it’s not obvious that you want that — in practice it’s more likely that you want the upper tail or the lower tail.

That said, at a fundamental level, the answer is the same: use the cumulative distribution function and/or its inverse, as appropriate.

Answer 2

Explanation:

The normal distribution is defined by its mean and standard deviation, so if we know that the heights of the seedlings are normally distributed and we know certain percentiles of the distribution, we can find the mean and standard deviation.

We are given that 10% of the seedlings are taller than 15 cm, which means that 90% of the seedlings are shorter than or equal to 15 cm. This tells us that the 90th percentile of the distribution is 15 cm.

Similarly, we are given that 5% of the seedlings are shorter than 4 cm, which means that 95% of the seedlings are taller than or equal to 4 cm. This tells us that the 95th percentile of the distribution is 4 cm.

Since the normal distribution is symmetrical, the mean of the distribution is the value that separates the upper and lower halves of the distribution. In this case, the mean is the value that separates the lower 95% from the upper 5%, which means it is the value that is exactly halfway between the 95th and 90th percentiles.

To find the mean, we can add the 95th percentile (4 cm) and the 90th percentile (15 cm) and divide by 2:

mean = (4 + 15) / 2

= 19 / 2

= 9.5 cm

To find the standard deviation, we can use the fact that the normal distribution has the property that approximately 68% of the values fall within one standard deviation of the mean, approximately 95% fall within two standard deviations, and approximately 99.7% fall within three standard deviations.

Since we know that the 95th percentile (4 cm) is 2 standard deviations below the mean and the 90th percentile (15 cm) is 2 standard deviations above the mean, we can set up the equation:

4 cm = mean - 2 * standard deviation

15 cm = mean + 2 * standard deviation

Solving for standard deviation, we get:

standard deviation = (15 cm - 4 cm) / (2 * 2)

= 11 cm / 4

= 2.75 cm

So, the mean height of the seedlings is 9.5 cm and the standard deviation is 2.75 cm.