The accompanying data represent the miles per gallon of a random sample of cars with a three-cylinder, 1.0 liter engine. (a) Compute the z-score corresponding to the individual who obtained 37.837.8 miles per gallon. Interpret this result. (b) Determine the quartiles. (c) Compute and interpret the interquartile range, IQR. (d) Determine the lower and upper fences. Are there any outliers? LOADING... Click the icon to view the data. (a) Compute the z-score corresponding to the individual who obtained 37.837.8 miles per gallon. Interpret this result. The z-score corresponding to the individual is nothing and indicates that the data value is nothing standard deviation(s) ▼ below above the ▼ mean. median. (Type integers or decimals rounded to two decimal places as needed.)

The question is not complete because we were not given the data that represent the miles per gallon of a random sample of smart cars with a three cylinder, 1.0-liter engine.

However, we are going to assume some data below such that whatever data is latter provided, the solution to the question follows the sama procedure and pattern.

Sample Data = 31.5, 36.0, 37.8, 38.5, 40.1, 42.2, 34.2, 36.2, 38.1, 38.7, 40.6, 42.5, 34.7, 37.3, 38.2, 39.5, 41.4, 43.4, 35.6, 37.6, 38.4, 39.6, 41.7, 49.3

Now to solve for

(a) which is to compute the Z-score corresponding to the individual who obtained 37.837.8 miles per gallon.

Therefore, the value of z-score is calculated as follows,

μ = Σ lin (n) lin (i=1) X₁ / n → 933.1 /24 → 38.88

σ = √ Σ lin (n) lin (i=1) X₁² - nX⁻²/ n - 1  → 3.61

z = X - μ / σ  → 37.8 - 38.88/3.61 = −  0.299

Therefore, the value of z-score is -0.299.

(b) Considering the fact that the values are in ascending order is given as,

Recalling the values: 31.5, 36, 37.8 ,38.5, 40.1 ,42.2, 34.2 ,36.2, 38.1, 38.7, 40.6 ,42.5, 34.7, 37.3 ,38.2, 39.5 ,41.4, 43.4, 35.6, 37.6 ,38.4 ,39.6 ,41.7, 49.3

Q₁ = ( n+1/4)th value → (24+1/4)th value → 6.25th value  →  36.2 + 0.25 ∗ ( 37.3 −  36.2 )  =>  36.47

M = ( n+1/2)th value → (24+1/2)th value = (24+1/4)th value =>  12.5 t h  v a l u e  => 38.4 + 38.5/2 = 38.45

Q₃ = 3( n+1/4)th value  → 3(24+1/4)th value =>  18.75 t h  v a l u e  =>  40.6 + 0.75 ∗  ( 41.4  −  40.6 )  =>  41.2

(c) I n t e r q u a r t i l e r a n g e  =  Q ₃ −  Q ₁  =>  41.2  −  36.47  =>  4.25

Therefore, the IQR value is measure of spread of data, and here the value of IQR = 4.25.

(d) L o w e r f o u r t h  =  Q₁− 1.5 ( I Q R )  =>  36.75 −  1.5  ( 4.25 )  =>  30.375  U p p e r f o u r t h  =  Q ₃ +  1.5  ( I Q R )  =>  41  +  1.5  ( 4.25 )  =>  47.375

Hence, there is outliers since observations are not less than 30.375 and greater than 47.375.

The table for the entire problem is given as,

Z- Score     -0.299

Q₁                  36.47

 Q₂                  38.45

 Q₃                  41.2

  IQR               4.25

  Lower fourth   30.375

  Upper fourth       47.375

step-by-step explanation:

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hi! i think the answer is 1/2 : )