A) For this problem, we will need to use a normal calculation, in that we find the z-score and the area to the right using Table A.
z = (10 - 7.65) / 1.45
z = 1.62
area to the left for a z-score of 1.62 = 0.9474
area to the right for a z-score of 1.62 = 0.0526
The probability that a randomly selected ornament will cost more than $10 is 0.0526 or 5.26%.
B) For this problem, we will use the binomial probability formula since the problem is asking for the probability that exactly 3 ornaments cost over $10. There are two forms of this equation. One is <em>nCr x p^r x q^n-r</em> and the other is <em>(n r) x p^r x (1 - p)^n-r</em>. I will show both formulas below.
8C3 x 0.0526^3 x 0.9474^5
(8 3) x 0.0526^3 x 0.9474^5
With both equations, the answer is the same. Whichever you are more familiar or comfortable with is the one I would recommend you use.
The probability that exactly 3 of the 8 ornaments cost over $10 is 0.00622 or 0.622%.
Hope this helps!! :)
Answer:
there answer is either a or c
Answer: 0.50, which is choice b
Explanation:
The interval 
covers 5-1 = 4 units in the horizontal direction.
This is out of 8 units that span from x = 0 to x = 8 (we could say 8-0 = 8).
So we get the final result of 4/8 = 0.50
In other words, the interval from x = 1 to x = 5 covers exactly half of the interval from x = 0 to x = 8.
5*8-(4+2)=34
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