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Sagot :
To solve the problem, we need to determine the probability that a randomly selected worker at the factory earns between [tex]$350 and $[/tex]450 per week, given that weekly wages are normally distributed with a mean of [tex]$400 and a standard deviation of $[/tex]50. We will use the 68%-95%-99.7% rule, also known as the empirical rule, which provides a quick way to estimate the probability for normally distributed data.
1. Identify the Mean and Standard Deviation:
- Mean ([tex]\(\mu\)[/tex]) = [tex]$400 - Standard Deviation (\(\sigma\)) = $[/tex]50
2. Calculate the Z-Scores:
The Z-score formula is given by:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(X\)[/tex] is the value for which we are calculating the Z-score, [tex]\(\mu\)[/tex] is the mean, and [tex]\(\sigma\)[/tex] is the standard deviation.
We need to calculate the Z-scores for the lower bound ([tex]$350) and the upper bound ($[/tex]450):
- For [tex]\(X = 350\)[/tex]:
[tex]\[ Z_{\text{lower}} = \frac{350 - 400}{50} = \frac{-50}{50} = -1.0 \][/tex]
- For [tex]\(X = 450\)[/tex]:
[tex]\[ Z_{\text{upper}} = \frac{450 - 400}{50} = \frac{50}{50} = 1.0 \][/tex]
So, we have the Z-scores:
[tex]\[ Z_{\text{lower}} = -1.0 \quad \text{and} \quad Z_{\text{upper}} = 1.0 \][/tex]
3. Apply the Empirical Rule:
According to the 68%-95%-99.7% rule for normal distributions, approximately:
- 68% of the data falls within one standard deviation ([tex]\(\sigma\)[/tex]) of the mean ([tex]\(\mu\)[/tex]), i.e., between [tex]\(\mu - \sigma\)[/tex] and [tex]\(\mu + \sigma\)[/tex].
Here, our Z-scores of [tex]\(-1\)[/tex] and [tex]\(1\)[/tex] correspond to one standard deviation below and above the mean, respectively. Therefore, the probability of a worker's wage being between [tex]$350 and $[/tex]450 is approximately 68%.
4. Conclusion:
The probability that a worker selected at random makes between [tex]$350 and $[/tex]450 per week is approximately 0.68 or 68%.
Thus, we have:
[tex]\[ Z_{\text{lower}} = -1.0, \quad Z_{\text{upper}} = 1.0, \quad \text{Probability} = 0.68 \][/tex]
1. Identify the Mean and Standard Deviation:
- Mean ([tex]\(\mu\)[/tex]) = [tex]$400 - Standard Deviation (\(\sigma\)) = $[/tex]50
2. Calculate the Z-Scores:
The Z-score formula is given by:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(X\)[/tex] is the value for which we are calculating the Z-score, [tex]\(\mu\)[/tex] is the mean, and [tex]\(\sigma\)[/tex] is the standard deviation.
We need to calculate the Z-scores for the lower bound ([tex]$350) and the upper bound ($[/tex]450):
- For [tex]\(X = 350\)[/tex]:
[tex]\[ Z_{\text{lower}} = \frac{350 - 400}{50} = \frac{-50}{50} = -1.0 \][/tex]
- For [tex]\(X = 450\)[/tex]:
[tex]\[ Z_{\text{upper}} = \frac{450 - 400}{50} = \frac{50}{50} = 1.0 \][/tex]
So, we have the Z-scores:
[tex]\[ Z_{\text{lower}} = -1.0 \quad \text{and} \quad Z_{\text{upper}} = 1.0 \][/tex]
3. Apply the Empirical Rule:
According to the 68%-95%-99.7% rule for normal distributions, approximately:
- 68% of the data falls within one standard deviation ([tex]\(\sigma\)[/tex]) of the mean ([tex]\(\mu\)[/tex]), i.e., between [tex]\(\mu - \sigma\)[/tex] and [tex]\(\mu + \sigma\)[/tex].
Here, our Z-scores of [tex]\(-1\)[/tex] and [tex]\(1\)[/tex] correspond to one standard deviation below and above the mean, respectively. Therefore, the probability of a worker's wage being between [tex]$350 and $[/tex]450 is approximately 68%.
4. Conclusion:
The probability that a worker selected at random makes between [tex]$350 and $[/tex]450 per week is approximately 0.68 or 68%.
Thus, we have:
[tex]\[ Z_{\text{lower}} = -1.0, \quad Z_{\text{upper}} = 1.0, \quad \text{Probability} = 0.68 \][/tex]
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