(-3C^4-5C^2-7C)+(4C^4+8C^3+2C^2) in standard form
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Answer:
a) 0.625; b) 16.879; c) 8.442
Step-by-step explanation:
Since this is a normal distribution, we want the value of the mean, μ: 2.2; and the value of the standard deviation, σ: 6.3.
For the 40th percentile, we look in a z chart. We want to find the value as close to 0.40 as we can get; this is 0.4013, and it corresponds to a z score of z = -0.25.
Our formula for a z score is . Using our values, we have
-0.25 = (X-2.2)/6.3
Multiply both sides by 6.3:
6.3(-0.25) = X-2.2
-1.575 = X-2.2
Add 2.2 to each side:
-1.575+2.2 = X-2.2+2.2
0.625 = X
For the 99th percentile, the value in the z chart closest to 0.99 is 0.9901, which corresponds to a z score of z = 2.33:
2.33 = (X-2.2)/6.3
Multiply both sides by 6.3:
6.3(2.33) = X-2.2
14.679 = X-2.2
Add 2.2 to each side:
14.679+2.2 = X-2.2+2.2
16.879 = X
For the IQR, we find the values for the 75th percentile (Q3) and the 25th percentile (Q1). The value in a z chart closest to 0.75 is 0.7486, which corresponds to a z score of z = 0.67:
0.67 = (X-2.2)/6.3
Multiply both sides by 6.3:
6.3(0.67) = X-2.2
4.221 = X-2.2
Add 2.2 to each side:
4.221+2.2 = X-2.2+2.2
6.421 = X
The value in a z chart closest to 0.25 is 0.2514, which corresponds to a z score of z = -0.67:
-0.67 = (X-2.2)/6.3
Multiply both sides by 6.3:
6.3(-0.67) = X-2.2
-4.221 = X-2.2
Add 2.2 to each side:
-4.221+2.2 = X-2.2+2.2
-2.021 = X
This makes the interquartile range
6.421--2.021 = 8.442