Question

Soft-drink cans are filled by an automated filling machine. The mean fill volume is 12.1 fluid ounces, and the standard deviation is 0.05 fluid ounce. Assume that the fill volumes of the cans are independent, normal random variables. What is the probability that the average volume of 10 cans selected from this process is less than 12 fluid ounces?

Answer 1

Answer:

Probability is 0.97725

Step-by-step explanation:

Given: Mean volume =12.1 ounces  ,standard deviation=0.05

  and 10 volumes are selected i.e no of observation or samples.

To find : calculate probability for fluid  less than the 12 ounces.i.e P(X<12) with normal distribution probability .

Solution:

Given that data is independent normal random variables , which depends upon mean ,variance and the Standard Z-score .

Hence we have ,

Mean=12.1

Variance =square of (standard deviation).

Normal distribution is given by ,

N(mean,variance)

As there are 10 cans with mean 12.1 and variance=(0.05)^2=0.025.

Normal distribution probability is given by sum of independent normal random variables.

Using the notation as P(X-Y>0)

I.e. P(X>Y)

finding the probability of X greater than Y i.e.(12.1>12) ounces .

with same standard deviation we get,

X-Y    ⇒N(12.1-12,0.0025)⇒N(0.1,0.0025).

=P(X-Y)

=P(X-Y>0)

=P(Z>$\frac{(0-0.1)}{\sqrt{0.0025} }$)

=P(Z>-2)

=P(Z<2).

=0.97725.

Using the Z- table refer the normal probability Z-table.