Given:
The given expression is
To find:
The two other equivalent expressions.
Solution:
We have,
Using distributive property, we get
So, one equivalent expression is .
On further simplification, we get
Therefore, the second and fully simplified equivalent expression is .
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.
Highest value in data - Lowest value in data = 60 - 28 = 32
(b) Interquartile rage = Third quartile - First quartile
=
=
observation in the data =
obs. { Because n = 9}
Therefore, = 42.
=
obs. = (3 x 2 ) =
obs = 50
So, Interquartile rage = 50 - 42 = 8.
2. (a) Sample mean formula = =
= 48.33
(b) Sample variance formula = where Xbar = Sample mean
= 92.75
(c) Sample standard deviation formula = =
= 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.
