QUESTION IMAGE
Question
- the average hours worked each week for the full - time teachers at southview high school is 55 hours per week and the standard deviation is 4.5 hours.
a) what is the maximum percentage of the teachers that might work an average of more than 64 hours each week?
b) if we assume the distribution is normal (or bell - shaped), what is the maximum percentage of the teachers that might work an average of more than 64 hours each week?
- bonus: if all the values of a data set must be within ( ssqrt{n - 1} ) of the mean, find the range that all the data values must be within for a data set with 50 data values, a mean of 60, and a standard deviation of 3.
Step1: Calculate the z - score
The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 64\), \(\mu=55\), and \(\sigma = 4.5\).
\(z=\frac{64 - 55}{4.5}=\frac{9}{4.5}=2\)
Step2: Use Chebyshev's theorem (for part a)
Chebyshev's theorem states that \(P(|X-\mu|\geq k\sigma)\leq\frac{1}{k^{2}}\). For \(k = 2\), \(P(X\geq\mu + k\sigma)\leq\frac{1}{k^{2}}\times\frac{1}{2}\) (because we are interested in the upper - tail).
\(P(X\geq55 + 2\times4.5)\leq\frac{1}{4}\times\frac{1}{2}=12.5\%\)
Step3: Use the standard normal distribution (for part b)
If \(X\sim N(\mu = 55,\sigma = 4.5)\), then for \(x = 64\), \(z = 2\).
Using the standard normal table \(P(Z>2)=1 - P(Z\leq2)\)
From the standard normal table \(P(Z\leq2)=0.9772\)
\(P(Z>2)=1 - 0.9772 = 0.0228=2.28\%\)
Step4: Solve the bonus problem
We are given \(n = 50\), \(\mu=60\), \(s = 3\)
The range is \(\mu\pm s\sqrt{n - 1}\)
First, calculate \(\sqrt{n - 1}=\sqrt{49}=7\)
The lower bound is \(60-3\times7=60 - 21 = 39\)
The upper bound is \(60 + 3\times7=60+21 = 81\)
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a) \(12.5\%\)
b) \(2.28\%\)
Bonus: The data values must be within the range \([39,81]\)