TO TRANSMIT INFORMATION ON THE INTERNET, LARGE FILES ARE BROKEN INTO PACKETS OF SMALLER SIZES. EACH PACKET HAS 1,500 BYTES OF IN
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A) 0.9803; 0.4803
B) 32
We calculate the z-score for this problem by using the formula:
Using our formula, we have:
Using a z-table (http://www.z-table.com) we see that the area to the left of, less than, this score is 0.9803.
To find the probability it is between the mean and this, we subtract the probability associated with the mean (0.5) from this:
0.9803 - 0.5 = 0.4803.
To find B, we first find the z-score for this. Using a z-table (http://www.z-table.com) we see that the closest z-score would be 2.33. We then set up our equation as
Multiplying both sides by 0.85 we have
2.33(0.85) = 0.35√n
1.9805 = 0.35√n
Divide both sides by 0.35:
1.9805/0.35 = √n
Square both sides:
(1.9805/0.35)² = n
32 ≈ n
B) 32
We calculate the z-score for this problem by using the formula:
Using our formula, we have:
Using a z-table (http://www.z-table.com) we see that the area to the left of, less than, this score is 0.9803.
To find the probability it is between the mean and this, we subtract the probability associated with the mean (0.5) from this:
0.9803 - 0.5 = 0.4803.
To find B, we first find the z-score for this. Using a z-table (http://www.z-table.com) we see that the closest z-score would be 2.33. We then set up our equation as
Multiplying both sides by 0.85 we have
2.33(0.85) = 0.35√n
1.9805 = 0.35√n
Divide both sides by 0.35:
1.9805/0.35 = √n
Square both sides:
(1.9805/0.35)² = n
32 ≈ n