Answer:
Step-by-step explanation:
The 90th percentile of a normally distributed curve occurs at 1.282 standard deviations. Similarly, the 10th percentile of a normally distributed curve occurs at -1.282 standard deviations.
To find the percentile for the television weights, use the formula:
, where is the average of the set, is some constant relevant to the percentile you're finding, and is one standard deviation.
As I mentioned previously, 90th percentile occurs at 1.282 standard deviations. The average of the set and one standard deviation is already given. Substitute , , and :
Therefore, the 90th percentile weight is 5.1282 pounds.
Repeat the process for calculating the 10th percentile weight:
The difference between these two weights is .
1
Add 1111 to both sides
5x\ge 31+115x≥31+11
2
Simplify 31+1131+11 to 4242
5x\ge 425x≥42
3
Divide both sides by 55
x\ge \frac{42}{5}x≥542
Expected to lose. they initially paid $350 for the tickets and are up charging the price. and the winning price is $100 less.
It should be 14/12 or 1 and 1/6