Question

The Blue Wild Fish Company sells a particular species of fish that live in a local lake. In the entire lake the distribution of fish lengths follows a normal distribution with mean 26 inches and standard deviation 7.8 inches.
You have 5 attempts for numeric answers and 1 attempt for multiple choice answers. Round answers to 2 decimal places.

1. A fisherman goes out and catches 45 fish every day. If we were to look at the distribution of the average lengths of the fish caught on each day we would be looking at a sampling distribution.
a. The mean of this sampling distribution would be ____ inches.
b. The standard deviation of this sampling distribution (called standard error) would be ____ inches,

2. How would mean and standard error of the sampling distribution change if took larger sample?
a. The mean would  --- increase decrease stay the same .
b. The standard error would  --- increase stay the same decrease .

3. Let's think about z-scores.
a. If the fisherman went out fishing and caught only one fish with a length of 40.7 inches, what would be the corresponding z-score?
b. If the fisherman went out fishing and caught the usual 45 fish and had a sample average length of 40.7 inches, what would be the corresponding z-score?
c. Is the fisherman more likely to catch one fish with a length of 40.7 inches or a group of 45 fish with an average length of 40.7 inches?  --- one fish 45 fish
d. If the fisherman went out fishing and caught the usual 45 fish and had a sample average length of 27.4 inches, what would be the corresponding z-score?
e. Is the fisherman more likely to catch one fish with a length of 40.7 inches or a group of 45 fish with an average length of 27.4 inches?  --- 45 fish one fish

4. Consider the following probabilities:
Put the smaller number then the larger number.
a. If the fisherman only catches one fish a day, then approximately the middle 99.7% of those days the fish will have a length between  and  inches.
b. If the fisherman catches the usual 45 fish a day, then approximately the middle 99.7% of those days the sample of fish will have an average length between  and  inches.

Answer 1

1)

a. The mean of this sampling distribution would be _26___ inches.
b. The standard deviation of this sampling distribution (called standard error) would be σ/√n = 7.8/√45 = 1.163  inches,

2)

a. The mean would  stay the same .
b. The standard error would   decrease .

3)

a)

Z =   (X - µ )/(σ/√n) = (   40.7   -   26.00   ) / (   7.800   / √   1   ) =   1.885

b)

Z =   (X - µ )/(σ/√n) = (   40.7   -   26.00   ) / (   7.800   / √   45   ) =   12.64

c)

the fisherman is more likely to catch a group of 45 fish with an average length of 40.7 inches

d)

Z =   (X - µ )/(σ/√n) = (   27.4   -   26.00   ) / (   7.800   / √   45   ) =   1.20

e)

the fisherman more likely to catch one fish with a length of 40.7 inches

4)

µ =    26                              
σ =    7.8                              
n=   1                              
proportion=   0.997                              
proportion left    0.003   is equally distributed both left and right side of normal curve                           
z value at   0.0015   = ±   -2.97   (excel formula =NORMSINV(   0.00   / 2 ) )          
z = ( x - µ ) / (σ/√n)                                  
so, X = z σ / √n + µ =                                  
X1 =   -2.97   *   7.8   / √   1   +   26   =   2.85
X2 =   2.97   *   7.8   / √   1   +   26   =   49.15
If the fisherman only catches one fish a day, then approximately the middle 99.7% of those days the fish will have a length between 2.85 and 49.15

b)

µ =    26                              
σ =    7.8                              
n=   45                              
proportion=   0.997                              
proportion left    0.003   is equally distributed both left and right side of normal curve                           
z value at   0.0015   = ±   -2.97   (excel formula =NORMSINV(   0.00   / 2 ) )          
z = ( x - µ ) / (σ/√n)                                  
so, X = z σ / √n + µ =                                  
X1 =   -2.97   *   7.8   / √   45   +   26   =   22.55
X2 =   2.97   *   7.8   / √   45   +   26   =   29.45
If the fisherman catches the usual 45 fish a day, then approximately the middle 99.7% of those days the sample of fish will have an average length between 22.55 and 29.45