Question

(1 point) The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 8.2 ounces and standard deviation 0.13 ounces.

(a) What is the probability that the average weight of a bar in a Simple Random Sample (SRS) with four of these chocolate bars is between 8.08 and 8.4 ounces?

(b) For a SRS of four of these chocolate bars, what is the level L such that there is a 3% chance that the average weight is less than L?

Answer 1

Let X be a random variable denoting the weights of chocolate bars produced by a certain machine.

X ~ N(8.2, 0.13)

where mean =
8.2
and standard deviation =
0= 1.3

(a) The average weight of a chocolate bar =

X

Now,

$P(8.08<\bar X<8.4 )=P(\frac{\sqrt 4 (8.08-8.2)}{0.13}<\frac{\sqrt n(\bar X-\mu)}{\sigma}<\frac{\sqrt 4(8.4-8.2)}{0.13})$

  

Where, $Z=\frac{\sqrt n(\bar X-\mu)}{\sigma}\sim N(0,1)$

$=\Phi (3.08)-\Phi (-1.85)$

  $=\Phi (3.08)-1+\Phi (1.85)=0.9989-1+0.9678$

= 0.9667

the probability that the average weight of a bar in a Simple Random Sample (SRS) with four of these chocolate bars is between 8.08 and 8.4 ounces is 0.9667.

We have to find the value of L such that,

$P(\bar X< L)=0.03$

$or,P(\bar X< L)=1-0.97$

$or,P(\bar X< L)=1-\Phi (1.89)=\Phi (-1.89)$

$or,P(\frac{\sqrt n (\bar X - \mu)}{\sigma}< \frac{\sqrt 4 (L - 8.2)}{0.13})=\Phi (-1.89)$

$or,P(Z< \frac{\sqrt 4 (L - 8.2)}{0.13})=\Phi (-1.89)$

Where, $Z=\frac{\sqrt n(\bar X-\mu)}{\sigma}\sim N(0,1)$

$or,\Phi (\frac {\sqrt 4 (L - 8.2)}{0.13})=\Phi (-1.89)$

$or,\frac {2 (L - 8.2)}{0.13}=-1.89$

= 8.077

the level L such that there is a 3% chance that the average weight is less than L is 8.077

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