Question

The Los Angeles Times regularly reports the air quality index for various areas of Southern California. A sample of air quality index values for Pomona provided the following data: 28, 42, 58, 48, 45, 55, 60, 49, and 50.

1. Compute the range and interquartile range.

a. Range
b. Interquartile range

2. Compute the sample mean, sample variance, and sample standard deviation (to 2 decimals).

a. Sample mean
b. Sample variance
c. Sample standard deviation

3. A sample of air quality index readings for Anaheim provided a sample mean of 48.3, a sample variance of 136, and a sample standard deviation of 11.66. What comparisons can you make between the air quality in Pomona and that in Anaheim on the basis of these descriptive statistics??

Answer 1

Answer:

1. (a) 32     (b) 8

2. (a) 48.33    (b) 92.75    (c) 9.63

3.  We observe that sample mean of air quality index in Pomona is higher that that of Anaheim which indicates that the level of air pollution is higher in Pomona and there is more health concern here.

On the other hand, the variation in air quality index is higher in Anaheim than in Pomona as Anaheim has higher variance and standard deviation for air quality index which means that there are not much fluctuations in air quality index in Pomona.  

Step-by-step explanation:

First arranging our data of air quality index values for Pomona in ascending order, we get : 28, 42, 45, 48, 49, 50, 55, 58, 60.

1. (a) Range is given by the formula :

                      Highest value in data - Lowest value in data = 60 - 28 = 32

        (b) Interquartile rage = Third quartile - First quartile

                                           = $Q_3 - Q_1$

             $Q_1$ =  $(\frac{n-1}{4} )^{th}$ observation in the data = $2^{nd}$ obs. { Because n = 9}

                 Therefore, $Q_1$ = 42.

             $Q_3$ = $(3Q_1)^{th}$ obs. = (3 x 2 ) = $6^{th}$ obs = 50

             So, Interquartile rage = 50 - 42 = 8.

     2. (a) Sample mean formula =  $\frac{\sum X_i}{n}$ = $\frac{28+ 42+ 45+ 48+ 49+ 50+ 55+ 58+ 60}{9}$ = 48.33

         (b) Sample variance formula = $\frac{\sum (X_i - Xbar)^{2}}{n-1}$ where Xbar = Sample mean

                                                         = 92.75

          (c) Sample standard deviation  formula = $\sqrt{variance}$ = $\sqrt{92.75}$ = 9.63.

     3. Given that a sample of air quality index readings for Anaheim has   sample mean of 48.3, a sample variance of 136, and a sample standard deviation of 11.66.

We observe that sample mean of air quality index in Pomona is higher that that of Anaheim which indicates that the level of air pollution is higher in Pomona and there is more health concern here.

On the other hand, the variation in air quality index is higher in Anaheim than in Pomona as Anaheim has higher variance and standard deviation for air quality index which means that there are not much fluctuations in air quality index in Pomona.  

                                       

Answer 2

(1)  a.The range will be =32    

     b. Interquarlite range = 8

2. (a) Sample mean=  48.33  

   (b) Sample variance= 92.75  

   (c) sample standard deviation= 9.63

3. It is observed that the sample mean of air quality index in Sonoma is higher than the anaheim so it is the indication of higher air pollution in Sonoma so there is more health concern here

On the other hand, the variation in air quality index is higher in Anaheim than in Pomona as Anaheim has higher variance and standard deviation for air quality index which means that there are not much fluctuations in the air quality index in Pomona.  

What will be the answers by analyzing the given data?

Here we have the air quality index values fpr ponoma  28, 42, 45, 48, 49, 50, 55, 58, 60.

Now first we will find the range of the data

(a) Range =  Highest value in data - Lowest value in data = 60 - 28 = 32

(b) Interquartile rage = Third quartile - First quartile

                                   = $Q_3 -Q_1$

$Q_1 =\dfrac{n-1}{4} th = \dfrac{9-1}{4} =2^{nd} term\\Q_1=42$

$Q_3=3Q_1 =3\times 2=6^{th} term\\Q_3 =50$                                          

So, Interquartile rage = 50 - 42 = 8.

2. (a) Sample mean formula =  $\dfrac{\sum X_i}{n}$  =  

   Mean  =    $\dfrac{28+42+45+48+49+50+55+58+60}{9} =48.33$

  (b) Sample variance formula =  where = $\overline {X}$ = Sample mean

                                $\rm Variance =\dfrac{ \sum(X_i -\overline X)^2}{n-1}=92.75$                      

 (c) Sample standard deviation  formula = $\sqrt{Variance } =9.63$

3. It is given that a sample of air quality index readings for Anaheim has   sample mean of 48.3, a sample variance of 136, and a sample standard deviation of 11.66.

It is observe that sample mean of air quality index in Pomona is higher that that of Anaheim which indicates that the level of air pollution is higher in Pomona and there is more health concern here.

On the other hand, the variation in air quality index is higher in Anaheim than in Pomona as Anaheim has higher variance and standard deviation for air quality index which means that there are not much fluctuations in air quality index in Pomona.  

Thus the above calculation gives all the answers for the air quality index of ponoma

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