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Sagot :
To solve this problem, we need to determine the probability that a pilot's weight falls between 150 lb and 201 lb, given that the weights of the pilots are normally distributed with a mean of 158 lb and a standard deviation of 33.9 lb.
Here is a detailed, step-by-step solution:
1. Identify the given values:
- Mean ([tex]\(\mu\)[/tex]) = 158 lb
- Standard deviation ([tex]\(\sigma\)[/tex]) = 33.9 lb
- Lower bound weight = 150 lb
- Upper bound weight = 201 lb
2. Calculate the z-scores for the lower and upper bounds:
The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is:
[tex]\[ z = \frac{(X - \mu)}{\sigma} \][/tex]
Let’s calculate the z-score for the lower bound (150 lb):
[tex]\[ z_{\text{lower}} = \frac{(150 - 158)}{33.9} = \frac{-8}{33.9} \approx -0.236 \][/tex]
Now, calculate the z-score for the upper bound (201 lb):
[tex]\[ z_{\text{upper}} = \frac{(201 - 158)}{33.9} = \frac{43}{33.9} \approx 1.268 \][/tex]
3. Determine the probability for the z-scores using the standard normal distribution:
To find the probability that a pilot's weight is between a lower z-score and an upper z-score, we use the cumulative distribution function (CDF) of the standard normal distribution. The CDF value for a given z-score represents the probability that a standard normal random variable is less than or equal to that z-score.
We need the probabilities associated with the two z-scores:
- Probability corresponding to [tex]\( z_{\text{lower}} = -0.236 \)[/tex]
- Probability corresponding to [tex]\( z_{\text{upper}} = 1.268 \)[/tex]
4. Fetch the standard normal distribution values for these z-scores:
- [tex]\( P(Z \leq -0.236) \approx 0.4072 \)[/tex]
- [tex]\( P(Z \leq 1.268) \approx 0.8982 \)[/tex]
5. Compute the probability that a pilot's weight is between the two bounds:
This is found by subtracting the cumulative probability at the lower z-score from the cumulative probability at the upper z-score.
[tex]\[ P(150 \leq X \leq 201) = P(Z \leq 1.268) - P(Z \leq -0.236) \approx 0.8982 - 0.4072 \approx 0.491 \][/tex]
Thus, the probability that a randomly selected pilot weighs between 150 lb and 201 lb is approximately [tex]\( 0.491 \)[/tex], rounded to four decimal places.
Here is a detailed, step-by-step solution:
1. Identify the given values:
- Mean ([tex]\(\mu\)[/tex]) = 158 lb
- Standard deviation ([tex]\(\sigma\)[/tex]) = 33.9 lb
- Lower bound weight = 150 lb
- Upper bound weight = 201 lb
2. Calculate the z-scores for the lower and upper bounds:
The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is:
[tex]\[ z = \frac{(X - \mu)}{\sigma} \][/tex]
Let’s calculate the z-score for the lower bound (150 lb):
[tex]\[ z_{\text{lower}} = \frac{(150 - 158)}{33.9} = \frac{-8}{33.9} \approx -0.236 \][/tex]
Now, calculate the z-score for the upper bound (201 lb):
[tex]\[ z_{\text{upper}} = \frac{(201 - 158)}{33.9} = \frac{43}{33.9} \approx 1.268 \][/tex]
3. Determine the probability for the z-scores using the standard normal distribution:
To find the probability that a pilot's weight is between a lower z-score and an upper z-score, we use the cumulative distribution function (CDF) of the standard normal distribution. The CDF value for a given z-score represents the probability that a standard normal random variable is less than or equal to that z-score.
We need the probabilities associated with the two z-scores:
- Probability corresponding to [tex]\( z_{\text{lower}} = -0.236 \)[/tex]
- Probability corresponding to [tex]\( z_{\text{upper}} = 1.268 \)[/tex]
4. Fetch the standard normal distribution values for these z-scores:
- [tex]\( P(Z \leq -0.236) \approx 0.4072 \)[/tex]
- [tex]\( P(Z \leq 1.268) \approx 0.8982 \)[/tex]
5. Compute the probability that a pilot's weight is between the two bounds:
This is found by subtracting the cumulative probability at the lower z-score from the cumulative probability at the upper z-score.
[tex]\[ P(150 \leq X \leq 201) = P(Z \leq 1.268) - P(Z \leq -0.236) \approx 0.8982 - 0.4072 \approx 0.491 \][/tex]
Thus, the probability that a randomly selected pilot weighs between 150 lb and 201 lb is approximately [tex]\( 0.491 \)[/tex], rounded to four decimal places.
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