Question

The weight of the contents of a can of Campbell's minestrone soup can be modelled by a normal distribution with a mean of 439.9 grams and a standard deviation of 3.2 grams. A random sample of 15 cans is selected for quality control testing. Determine the probability that (4 decimal places)

1. A randomly selected can weighs more than 437.5 grams,

2. The sample mean of the cans taken for quality control testing will NOT be within the target range of 438.0 to 442.0 grams.

Answer 1

Solution-1

X~(439.9 ,3.2)

n=15

s=sigma/sqrt(1)=3.2/sqrt(1)= 3.2

xbar~N(439.9,3.2)

P(xbar>437.5)

P(z>437.5-439.9/3.2)

=P(z>-0.75)

=1-P(Z<0.75)

=1-0.2266

=0.7734

0.7734

Solution-2:

s=sigma/sqrt(15)=3.2/sqrt(15)= 0.8262364

xbar~N(439.9,0.8262364)

P(438<xbar<442)

=P(438-439.9/0.8262364<Z<442-439.9/0.8262364)

=P(0.6052<z<5.4464)

=P(Z<5.4464)-P(Z<0.6052)

=1-0.7275

=0.2725

will NOT be within the target range=1-0.2725=0.7275

0.7275