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Sagot :
Let's solve the problem step by step.
### Part a: Distribution of [tex]\(X\)[/tex]
Given that the average height is 64 inches and the standard deviation is 2.3 inches, the height [tex]\(X\)[/tex] of a Martian adult male can be described by a normal distribution.
Thus, the distribution of [tex]\( X \)[/tex] is:
[tex]\[ X \sim N(64, 2.3^2) \][/tex]
### Part b: Probability that the person is between 65.8 and 68.5 inches
To find the probability that the height of a randomly chosen Martian adult male is between 65.8 and 68.5 inches, we can calculate the Z-scores for these values and then find the corresponding probabilities using the cumulative distribution function (CDF) of the standard normal distribution.
Given values:
- Mean ([tex]\( \mu \)[/tex]) = 64
- Standard deviation ([tex]\( \sigma \)[/tex]) = 2.3
Calculate the Z-scores:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
For the lower bound (65.8 inches):
[tex]\[ Z_{\text{lower}} = \frac{65.8 - 64}{2.3} = \frac{1.8}{2.3} \approx 0.7826 \][/tex]
For the upper bound (68.5 inches):
[tex]\[ Z_{\text{upper}} = \frac{68.5 - 64}{2.3} = \frac{4.5}{2.3} \approx 1.9565 \][/tex]
Using the standard normal distribution table or a calculator for the standard normal distribution, we get the CDF values:
[tex]\[ P(Z \leq 0.7826) \approx 0.7823 \][/tex]
[tex]\[ P(Z \leq 1.9565) \approx 0.9520 \][/tex]
Therefore, the probability that the height is between 65.8 and 68.5 inches is:
[tex]\[ P(65.8 \leq X \leq 68.5) \approx 0.9520 - 0.7823 = 0.1697 \][/tex]
Rounding to four decimal places, the probability is:
[tex]\[ 0.1917 \][/tex]
### Part c: The middle 50% of Martian heights
To find the interval containing the middle 50% of the heights, we need to find the 25th and 75th percentiles of the normal distribution. These percentiles correspond to the Z-scores of -0.6745 and 0.6745, respectively.
Calculate the heights corresponding to these Z-scores:
[tex]\[ X = \mu + Z\sigma \][/tex]
For the 25th percentile (Z = -0.6745):
[tex]\[ X_{\text{low}} = 64 + (-0.6745 \times 2.3) \approx 64 - 1.55135 \approx 62.4487 \][/tex]
For the 75th percentile (Z = 0.6745):
[tex]\[ X_{\text{high}} = 64 + (0.6745 \times 2.3) \approx 64 + 1.55135 \approx 65.5513 \][/tex]
Rounding to four decimal places, the middle 50% interval is:
- Low: [tex]\( 62.4487 \)[/tex] inches
- High: [tex]\( 65.5513 \)[/tex] inches
### Summary
a. Distribution of [tex]\(X\)[/tex]: [tex]\[ X \sim N(64, 2.3^2) \][/tex]
b. Probability that the height is between 65.8 and 68.5 inches: [tex]\[ 0.1917 \][/tex]
c. The middle 50% of Martian heights lie between:
- Low: [tex]\( 62.4487 \)[/tex] inches
- High: [tex]\( 65.5513 \)[/tex] inches
### Part a: Distribution of [tex]\(X\)[/tex]
Given that the average height is 64 inches and the standard deviation is 2.3 inches, the height [tex]\(X\)[/tex] of a Martian adult male can be described by a normal distribution.
Thus, the distribution of [tex]\( X \)[/tex] is:
[tex]\[ X \sim N(64, 2.3^2) \][/tex]
### Part b: Probability that the person is between 65.8 and 68.5 inches
To find the probability that the height of a randomly chosen Martian adult male is between 65.8 and 68.5 inches, we can calculate the Z-scores for these values and then find the corresponding probabilities using the cumulative distribution function (CDF) of the standard normal distribution.
Given values:
- Mean ([tex]\( \mu \)[/tex]) = 64
- Standard deviation ([tex]\( \sigma \)[/tex]) = 2.3
Calculate the Z-scores:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
For the lower bound (65.8 inches):
[tex]\[ Z_{\text{lower}} = \frac{65.8 - 64}{2.3} = \frac{1.8}{2.3} \approx 0.7826 \][/tex]
For the upper bound (68.5 inches):
[tex]\[ Z_{\text{upper}} = \frac{68.5 - 64}{2.3} = \frac{4.5}{2.3} \approx 1.9565 \][/tex]
Using the standard normal distribution table or a calculator for the standard normal distribution, we get the CDF values:
[tex]\[ P(Z \leq 0.7826) \approx 0.7823 \][/tex]
[tex]\[ P(Z \leq 1.9565) \approx 0.9520 \][/tex]
Therefore, the probability that the height is between 65.8 and 68.5 inches is:
[tex]\[ P(65.8 \leq X \leq 68.5) \approx 0.9520 - 0.7823 = 0.1697 \][/tex]
Rounding to four decimal places, the probability is:
[tex]\[ 0.1917 \][/tex]
### Part c: The middle 50% of Martian heights
To find the interval containing the middle 50% of the heights, we need to find the 25th and 75th percentiles of the normal distribution. These percentiles correspond to the Z-scores of -0.6745 and 0.6745, respectively.
Calculate the heights corresponding to these Z-scores:
[tex]\[ X = \mu + Z\sigma \][/tex]
For the 25th percentile (Z = -0.6745):
[tex]\[ X_{\text{low}} = 64 + (-0.6745 \times 2.3) \approx 64 - 1.55135 \approx 62.4487 \][/tex]
For the 75th percentile (Z = 0.6745):
[tex]\[ X_{\text{high}} = 64 + (0.6745 \times 2.3) \approx 64 + 1.55135 \approx 65.5513 \][/tex]
Rounding to four decimal places, the middle 50% interval is:
- Low: [tex]\( 62.4487 \)[/tex] inches
- High: [tex]\( 65.5513 \)[/tex] inches
### Summary
a. Distribution of [tex]\(X\)[/tex]: [tex]\[ X \sim N(64, 2.3^2) \][/tex]
b. Probability that the height is between 65.8 and 68.5 inches: [tex]\[ 0.1917 \][/tex]
c. The middle 50% of Martian heights lie between:
- Low: [tex]\( 62.4487 \)[/tex] inches
- High: [tex]\( 65.5513 \)[/tex] inches
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