. The owner of a fish market has an assistant who has determined that the weight

Question

. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of µ= 3.2 pounds and a standard deviation of  0.8 pounds.

(i)If a sample of 16 fish is taken what is the probability that the sample mean will be between 2.6 and 4.0 pounds? That is , find             

P(2.6 less than or equal to MEAN less than or equal to 4.0)

(ii) If a sample of 9 fish is taken what is the probability that the sample mean will be more than 4 pounds? That is , find              

P(Mean is greater than 4.0)

(iii) Suppose the population standard deviation is unknown. If 60% of all sample means are greater than 3 pounds and the population mean is still 3.2 pounds, what is the value of the population standard deviation?

Explanation / Answer

i)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    2.6      
x2 = upper bound =    4      
u = mean =    3.2      
n = sample size =    16      
s = standard deviation =    0.8      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -3      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    4      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.001349898      
P(z < z2) =    0.999968329      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.998618431   [ANSWER]

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

ii)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    4      
u = mean =    3.2      
n = sample size =    9      
s = standard deviation =    0.8      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    3      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3   ) =    0.001349898 [ANSWER]

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

iii)

The z score corresponding to a 0.60 right tailed area, by table/technology, is

z = -0.253347103

Thus, as

z = (x-u)*sqrt(n)/s

Then

-0.253347103 = (3-3.2)*sqrt(9)/s

Then

s = 2.368292325 [ANSWER]

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