Question

7. Suppose the amount of a popular sport drink in bottles leaving the filling machine has approximately a normal distribution with mean 101.5 milliliters (ml) and standard deviation 1.6 ml.

1. If the bottles are labeled 100 ml, what proportion of the bottles contain less than the labeled amount?

2. If only 2.5% of the bottles have contents that exceed a specified amount c, what is the value of c?

Answer 1

Let X be the amount of a popular sport drink in bottles leaving the filling machine.

Now, as X follows normal distribution with mean of 101.5 ml and standard deviation of 1.6 ml.

a)

For X = 100 ml, Z score = = z = (X - μ) / σ = (100 - 101.5) / 1.6 = -0.9375

So, P (X < 100) = P (z< -0.9375) = 0.1762 - (0.1762-0.1736) / 0.01 * 0.0075 = 0.17425 by interpolation using z distribution calculator (refer z table)

So, 17.425 % of bottles contain less than the labelled amount of drink.

b)

For top 2.5% amount, P(X > c) = 2.5% = 0.025 which gives a less than probability of 97.5% or 0.975

Looking for P = 0.975 in z table, Z score = 1.96

So, raw score = c = mean + z * standard deviation = 101.5 + 1.96*1.6 = 104.636 ml

So, only 2.5% of the bottles have contents that exceed a specified amount of 104.636 ml