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Answer :
Using the Z-score formula and a standard normal distribution table, the probability that a randomly selected pilot's weight is between 120 lb and 181 lb, given a normal weight distribution with a mean of 128 lb and a standard deviation of 32.4 lb, is approximately 0.5469 or 54.69%.
To calculate the probability that a randomly selected pilot's weight is between 120 lb and 181 lb, given a normal distribution with a mean of 128 lb and a standard deviation of 32.4 lb, we will use the Z-score formula. The Z-score is given by Z = (X - mean) / (standard deviation), where X is the value for which we are finding the Z-score.
For X = 120 lb (the lower limit),
Z1 = (120 - 128) / 32.4 = -0.2469 (approximately).
For X = 181 lb (the upper limit),
Z2 = (181 - 128) / 32.4 = 1.6358 (approximately).
Using a standard normal distribution table or a calculator, we can find the area (probability) to the left of Z1 and the area to the left of Z2. Subtracting these gives us the probability that the weight is between 120 lb and 181 lb:
[tex]P(Z1 < Z < Z2) = P(Z < Z2) - P(Z < Z1)[/tex]
By looking up Z1 and Z2 in a Z-table or using an online calculator:
- P(Z < -0.2469)
= 0.4026 (approximately) - P(Z < 1.6358)
= 0.9495 (approximately)
Therefore, the probability that a randomly selected pilot weighs between 120 lb and 181 lb is:
0.9495 - 0.4026 = 0.5469
Rounded to four decimal places, the probability is 0.5469, or approximately 54.69%.
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