Question 1168277: In New York State, the mean salary for high school teachers in 2017 was $98,210 with a standard deviation of $9,560. Only Alaska’s mean salary was higher! Assume New York’s state salaries follow a normal distribution.
a. What percent of New York’s state high school teachers earn between $84,000 and $89,000? (Round intermediate calculations to 2 decimal places and final answer to 2 decimal places.)
b. What percent of New York’s state high school teachers earn between $89,000 and $104,000? (Round intermediate calculations to 2 decimal places and final answer to 2 decimal places.)
c. What percent of New York’s state high school teachers earn less than $74,000? (Round intermediate calculations to 2 decimal places and final answer to 2 decimal places.)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let $X$ be the random variable representing the salary of a high school teacher in New York State in 2017. We are given that $X$ follows a normal distribution with a mean ($\mu$) of $98,210 and a standard deviation ($\sigma$) of $9,560.
We need to calculate the percentages for different salary ranges by first converting the salary values to z-scores using the formula:
$z = \frac{x - \mu}{\sigma}$
**a. What percent of New York’s state high school teachers earn between $84,000 and $89,000?**
First, calculate the z-scores for $x_1 = 84,000$ and $x_2 = 89,000$:
$z_1 = \frac{84,000 - 98,210}{9,560} = \frac{-14,210}{9,560} \approx -1.49$
$z_2 = \frac{89,000 - 98,210}{9,560} = \frac{-9,210}{9,560} \approx -0.96$
Now, we need to find the area under the standard normal curve between $z_1 = -1.49$ and $z_2 = -0.96$. We can find the cumulative probabilities for each z-score from a standard normal distribution table or using a calculator.
$P(Z < -1.49) \approx 0.0681$
$P(Z < -0.96) \approx 0.1685$
The percentage of teachers earning between $84,000 and $89,000 is:
$P(-1.49 < Z < -0.96) = P(Z < -0.96) - P(Z < -1.49) = 0.1685 - 0.0681 = 0.1004$
Rounding the final answer to 2 decimal places, the percentage is $10.04\%$.
**b. What percent of New York’s state high school teachers earn between $89,000 and $104,000?**
First, calculate the z-scores for $x_1 = 89,000$ (which we already calculated as $z_2 \approx -0.96$) and $x_3 = 104,000$:
$z_3 = \frac{104,000 - 98,210}{9,560} = \frac{5,790}{9,560} \approx 0.61$
Now, we need to find the area under the standard normal curve between $z_2 = -0.96$ and $z_3 = 0.61$. We can find the cumulative probabilities for each z-score from a standard normal distribution table or using a calculator.
$P(Z < -0.96) \approx 0.1685$
$P(Z < 0.61) \approx 0.7291$
The percentage of teachers earning between $89,000 and $104,000 is:
$P(-0.96 < Z < 0.61) = P(Z < 0.61) - P(Z < -0.96) = 0.7291 - 0.1685 = 0.5606$
Rounding the final answer to 2 decimal places, the percentage is $56.06\%$.
**c. What percent of New York’s state high school teachers earn less than $74,000?**
First, calculate the z-score for $x_4 = 74,000$:
$z_4 = \frac{74,000 - 98,210}{9,560} = \frac{-24,210}{9,560} \approx -2.53$
Now, we need to find the area under the standard normal curve to the left of $z_4 = -2.53$. We can find the cumulative probability for this z-score from a standard normal distribution table or using a calculator.
$P(Z < -2.53) \approx 0.0057$
Rounding the final answer to 2 decimal places, the percentage is $0.57\%$.
Final Answer: The final answer is $\boxed{a) 10.04\%, b) 56.06\%, c) 0.57\%}$
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