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

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