I measure the length of my bed 50 times each time making a r

I measure the length of my bed 50 times, each time making a random error, uniformly distributed over (-1,1) centimeters. What is the probability that the average of these 50 measurements differs from the actual value by less than 0.25 centimeters?

Solution

I measure the length of my bed 50 times, each time making a random error, uniformly distributed over (-1,1) centimeters.

What is the probability that the average of these 50 measurements differs from the actual value by less than 0.25 centimeters?

Let, Xi be the error in the ith measurement, then Xi is Uniform over (-1,1).

So E(Xi) = (-1 + 1) / 2 = 0

and var(Xi) = (1 - (-1)² ) / 12 = 1/3

Let, X = X1+X2+......+X50

then, E(X) = 50*0 = 0

Var(X) = var(Xi) = 50/3

By the CLT,

X is approximately normally distributed with mean 0 and variance 50/3.

Now X/50 is the average error of the 50 measurements, so we want to estimate the probability that X/50 is between -0.25 and 0.25

P(-0.25 < X/50 < 0.25) = P( -12.5 / sqrt(50/3) < X/ sqrt(50/3) < 12.5 / sqrt(50/3) )

=P(-3.06 < Z < 3.06)

P(Z < 3.06) - P(Z < -3.06)

EXCEL syntax for finding this probability is,

=NORMSDIST(z)

where, z is test statistic value.

P(Z < 3.06) = 0.9989

P(Z < -3.06) = 0.0011

P(-3.06 < Z < 3.06) = 0.9989 - 0.0011 = 0.9978