Question

The volume of a shampoo illed into a container is uniformly distributed between 374 and 380 milliliters. (a) What are the mean and standard deviation of the volume of shampoo? (b) What is the probability that the container is illed with less than the advertised target of 375 milliliters? (c) What is the volume of shampoo that is exceeded by 95% of the containers? (d) Every milliliter of shampoo costs the producer $0.002. Any shampoo more than 375 milliliters in the container is an extra cost to the producer. What is the mean extra cost?

Answer 1

Answer:

a) Mean = 377 mm

S.D = 1.73 mm

b) 16.7%

c) 374.3 mm

d) $0.0043 per container.

Step-by-step explanation:

a) The mean of a uniform distribution for a minimum value of 374 and a maximum value of 380 is:

$E(X)=\dfrac{Min+Max}{2}=\dfrac{374+380}{2}=377$

The standard deviation can be calculated as:

$\sigma=\dfrac{Max-Min}{\sqrt{12}}=\dfrac{380-374}{3.464}=1.73$

b) As 375 is within our maximum and minimum value for the distribution, we can calculate this probability as:

$P(X<375)=\dfrac{375-Min}{Max-Min}=\dfrac{375-374}{380-374}=\dfrac{1}{6}\approx 0.167$

c) We have to calculate Y so that P(X>Y)=0.95. The probability for Y can be written as:

$P(X>Y)=\dfrac{Max-Y}{Max-Min}=\dfrac{380-Y}{6}=0.95\\\\\\380-Y=0.95*6=5.7\\\\Y=380-5.7=374.3$

d) As we calculated in point (b), 16.7% of the containers have less than 375 milimiters.

Then, (100-16.7)%=83.3% of the containers have more than 375 milimieters.

The mean volume for the containers that have more than 375 mm can be easily calculated because all the values have constant and equal probability within the minimum (in this case, 375 mm) and maximum value (380 mm).

$E(Z_{375})=\dfrac{375+380}{2}=377.5$

This mean is 2.5 mm over 375 mm, so we have an extra cost of $0.002*2.5=$0.005 per container.

If 83.3% of the containers have an average of 2.5 mm of extra volumen, we can calculate the mean extra cost per container as:

$C=\%P\cdot \DeltaV=0.833\cdot0.005=0.004165\approx0.0042$