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Let ' $x$ ' be a random variable that represents the length of time it takes a student to write a term paper for Dr. Adam's Sociology class. After interviewing many students, it was found that ' $x$ ' has an approximately normal distribution with a mean of $\mu=7.3$ hours and standard deviation of $\alpha=0.8$ hours.
For parts a, b, c, Convert each of the following $x$ intervals to standardized $z$ intervals.
a.) $x<8.3$
b.) $x>9.9$
c.) $5.79.9)=$
f.) $P(5.7
Solution
Step 1 ：To convert the x intervals to standardized z intervals, we use the formula: \(z = \frac{x - \mu}{\alpha}\) where \(\mu\) is the mean and \(\alpha\) is the standard deviation.
Step 2 ：For \(x < 8.3\), we substitute \(x = 8.3\), \(\mu = 7.3\), and \(\alpha = 0.8\) into the formula to get: \(z = \frac{8.3 - 7.3}{0.8} = 1.25\). So, the z interval is \(z < 1.25\).
Step 3 ：For \(x > 9.9\), we substitute \(x = 9.9\), \(\mu = 7.3\), and \(\alpha = 0.8\) into the formula to get: \(z = \frac{9.9 - 7.3}{0.8} = 3.25\). So, the z interval is \(z > 3.25\).
Step 4 ：For \(5.7 < x < 8.5\), we substitute \(x = 5.7\) and \(x = 8.5\), \(\mu = 7.3\), and \(\alpha = 0.8\) into the formula to get: \(z1 = \frac{5.7 - 7.3}{0.8} = -2\) and \(z2 = \frac{8.5 - 7.3}{0.8} = 1.5\). So, the z interval is \(-2 < z < 1.5\).
Step 5 ：For \(P(x < 8.3)\), we look up the value for \(z = 1.25\) in the standard normal distribution table. The value is 0.8944. So, \(P(x < 8.3) = 0.8944\).
Step 6 ：For \(P(x > 9.9)\), we look up the value for \(z = 3.25\) in the standard normal distribution table. The value is 0.9994. However, since we are looking for the probability that x is greater than 9.9, we subtract this value from 1. So, \(P(x > 9.9) = 1 - 0.9994 = 0.0006\).
Step 7 ：For \(P(5.7 < x < 8.5)\), we look up the values for \(z = -2\) and \(z = 1.5\) in the standard normal distribution table. The values are 0.0228 and 0.9332, respectively. We subtract the smaller value from the larger value to get the probability. So, \(P(5.7 < x < 8.5) = 0.9332 - 0.0228 = 0.9104\).
Step 8 ：The final answers are: \(\boxed{P(x < 8.3) = 0.8944}\), \(\boxed{P(x > 9.9) = 0.0006}\), and \(\boxed{P(5.7 < x < 8.5) = 0.9104}\).
From Solvely APP
Solution
Step 1 ：To convert the x intervals to standardized z intervals, we use the formula: \(z = \frac{x - \mu}{\alpha}\) where \(\mu\) is the mean and \(\alpha\) is the standard deviation.
Step 2 ：For \(x < 8.3\), we substitute \(x = 8.3\), \(\mu = 7.3\), and \(\alpha = 0.8\) into the formula to get: \(z = \frac{8.3 - 7.3}{0.8} = 1.25\). So, the z interval is \(z < 1.25\).
Step 3 ：For \(x > 9.9\), we substitute \(x = 9.9\), \(\mu = 7.3\), and \(\alpha = 0.8\) into the formula to get: \(z = \frac{9.9 - 7.3}{0.8} = 3.25\). So, the z interval is \(z > 3.25\).
Step 4 ：For \(5.7 < x < 8.5\), we substitute \(x = 5.7\) and \(x = 8.5\), \(\mu = 7.3\), and \(\alpha = 0.8\) into the formula to get: \(z1 = \frac{5.7 - 7.3}{0.8} = -2\) and \(z2 = \frac{8.5 - 7.3}{0.8} = 1.5\). So, the z interval is \(-2 < z < 1.5\).
Step 5 ：For \(P(x < 8.3)\), we look up the value for \(z = 1.25\) in the standard normal distribution table. The value is 0.8944. So, \(P(x < 8.3) = 0.8944\).
Step 6 ：For \(P(x > 9.9)\), we look up the value for \(z = 3.25\) in the standard normal distribution table. The value is 0.9994. However, since we are looking for the probability that x is greater than 9.9, we subtract this value from 1. So, \(P(x > 9.9) = 1 - 0.9994 = 0.0006\).
Step 7 ：For \(P(5.7 < x < 8.5)\), we look up the values for \(z = -2\) and \(z = 1.5\) in the standard normal distribution table. The values are 0.0228 and 0.9332, respectively. We subtract the smaller value from the larger value to get the probability. So, \(P(5.7 < x < 8.5) = 0.9332 - 0.0228 = 0.9104\).
Step 8 ：The final answers are: \(\boxed{P(x < 8.3) = 0.8944}\), \(\boxed{P(x > 9.9) = 0.0006}\), and \(\boxed{P(5.7 < x < 8.5) = 0.9104}\).
