Tom K. answered 11/01/20
Knowledgeable and Friendly Math and Statistics Tutor
We use the Central Limit Theorem. Since the Uniform distribution has finite variance, and we have independent samples, it meets the CLT conditions.
For X ~ U[2, 3.7], µ = (2 + 3.7)/2 = 2.85 or 2 17/20, var = (3.7 - 2)2/12 = 289/1200
Then, for the sample with n=12, µ = 2.85 and var = 289/1200/42 = 289/50400, and σ = √var = 17√14/840 =
0.0757240185418536
The Central Limit Theorem tells us that the distribution approaches the normal distribution. That the original distribution is the uniform and n is relatively large assures us that it will be "close", and we are looking at quartiles, so all should be well.
Then, the distribution of the sample is approximately N(2 17/20, 289/50400) or
N(2.85, 0.00573412698412698)
Then, for the first quartile, x.25 = µ + z.25 σ = 2.85 + normsinv(.25) * 0.0757240185418536 =
2.79892492564986
From symmetry of the normal, The IQR (interquartile range) equals 2 * z.75 σ = 2 * normsinv(.75) * 0.0757240185418536 = 0.102150148700277