A variety of packaging solutions exist for products that must be kept within a s

Question

A variety of packaging solutions exist for products that must be kept within a specific temperature range. A cold chain distribution is a temperature-controlled supply chain. An unbroken cold chain is an uninterrupted series of storage and distribution activities that maintain a given temperature range. Cold chains are particularly useful in the food and pharmaceutical industries. A common suggested temperature range for a cold chain distribution in pharmaceutical industries is between 2 and 8 degrees Celsius. Gopal Vasudeva works in the packaging branch of Merck & Co. He is in charge of analyzing a new package that the company has developed. With repeated trials. Gopal has determined that the mean temperature that this package is able to maintain during its use is 5.68C with a standard deviation of 1.2 degree C. Assume that the temperature is normally distributed. In a report, use the sample information to: Calculate and interpret the probability that temperature goes (a) below 2 degree C and (b) above 8 degree C. Calculate and interpret the 5th and the 95th percentiles of the temperature that the package maintains.

Explanation / Answer

1.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    2      
x2 = upper bound =    8      
u = mean =    5.68      
          
s = standard deviation =    1.2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -3.066666667      
z2 = upper z score = (x2 - u) / s =    1.933333333      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.0010823      
P(z < z2) =    0.973402426      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.972320125   [ANSWER]

Hence, 97.23% of cold chains have temperatures between 2 and 8 degrees Celcius, which satisfies the common temperature range.

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2.

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.05      
          
Then, using table or technology,          
          
z =    -1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    5.68      
z = the critical z score =    -1.644853627      
s = standard deviation =    1.2      
          
Then          
          
x = 5TH PERCENTILE =    3.706175648   [ANSWER]

Hence, only 5% of cold chains have temperatures less than 3.706 degrees Celcius.

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First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    5.68      
z = the critical z score =    1.644853627      
s = standard deviation =    1.2      
          
Then          
          
x = 95TH PERCENTILE =    7.653824352   [ANSWER]     

Hence, only 5% of cold chains have temperatures more than 7.654 degrees Celcius.

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