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The weights of adult Fox Terriers in US are normally distributed, with a mean of

ID: 3126374 • Letter: T

Question

The weights of adult Fox Terriers in US are normally distributed, with a mean of 15 pounds and a standard deviation of 3 pounds The probability that the randomly chosen Fox Terrier weighs between 9 pounds and 18 pounds is closest to 0.7961 0.2514 0.8185 0.0475 .9725 To be in the top 13% of the weights, a Fox Terrier should weigh at least 18.39 pounds 25.27 pounds 17.52 pounds 20.64 pounds 15 pounds A random sample of 100 Fox Terriersis drawn fom this population. Identify the mean mu_x and standard error sigma_x of the sample mean weight x mu_x=15, sigma_x = .3 x mu_x=1.5, sigma_x = 3 x mu_x=15, sigma_x = 3 x mu_x=15, sigma_x = 30 x mu_x=1.5, sigma_x = .3 Find the probability that the sample mean weight exceeds 16 pounds .996 .5000 .6293 .3707 .004

Explanation / Answer

13.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    9      
x2 = upper bound =    18      
u = mean =    15      
          
s = standard deviation =    3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.818594614 [ANSWER, C]

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14.

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.13 =    0.87      
          
Then, using table or technology,          
          
z =    1.126391129      
          
As x = u + z * s,          
          
where          
          
u = mean =    15      
z = the critical z score =    1.126391129      
s = standard deviation =    3      
          
Then          
          
x = critical value =    18.37917339   [ANSWER, A]  

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15.

The mean remains the same, but the standard deviation is

sigma(X) = sigma/sqrt(n) = 3/sqrt(100) = 0.3

Hence,

OPTION A: ux = 15, sigma(X) = 0.3 [ANSWER, A]

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16.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    16      
u = mean =    15      
n = sample size =    100      
s = standard deviation =    3      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    3.333333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3.333333333   ) =    0.00042906 [ANSWER, E]