Suppose packages of cream cheese, coming from an automated processor, have weigh

Question

Suppose packages of cream cheese, coming from an automated processor, have weights that are normally distributed. For one day’s production run, the mean is 8.2 ounces and standard deviation is 0.1 ounce.

(a) If the packages of cream cheese are labeled 8 ounces, what proportion of the packages weigh less than the labeled amount?

(b) If a package of cream cheese is selected at random, what is the probability that this package weighs greater than 8.26 ounces?

(c) If only 5% of the packages exceed a specified weight w, what is the value of w?

(d) Suppose 4 packages are selected at random from the day’s production. What is the probability that the average weight of the four packages is greater than 8.26 ounces?

(e) A foreign company ordered 30 packages of cream cheese. There is a weight limit on international shipping. The maximum weight for international shipping is 245 ounces. Find the probability that this foreign company will receive their packages of cream cheese.

(a) 0.0228 (b) 0.2743 (c) 8.3645 (d) 0.1151 (e) 0.0336 (or 0.0475)

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    8      
u = mean =    8.2      
          
s = standard deviation =    0.1      
          
Thus,          
          
z = (x - u) / s =    -2      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2   ) =    0.022750132 [ANSWER]

****************************

b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    8.26      
u = mean =    8.2      
          
s = standard deviation =    0.1      
          
Thus,          
          
z = (x - u) / s =    0.6      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.6   ) =    0.274253118 [ANSWER]

*************************

c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.05 =   0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    8.2      
z = the critical z score =    1.644853627      
s = standard deviation =    0.1      
          
Then          
          
x = critical value =    8.364485363 oz [ANSWER]

******************************

d)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    8.26      
u = mean =    8.2      
n = sample size =    4      
s = standard deviation =    0.1      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.2   ) =    0.11506967 [ANSWER]

******************************

e)

That means the mean should be less than 245/30 = 8.16666667.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    8.166666667      
u = mean =    8.2      
n = sample size =    30      
s = standard deviation =    0.1      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.825741858 = -1.83      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.83   ) =    0.0336 [ANSWER]

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.