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Communication and Thinking The time it takes to drive from Orangeville to the Vaughan Mills Mall is normally distributed with a mean of 52 minutes and standard deviation of 5 minutes. What intervals could you estimate, using the knowledge from this activity, that do not include the mean as a max or min? For example, P of X is less than or equal to 52 is 50% is an interval we could estimate. However, it includes the mean as the maximum of the interval. Another example is P of 52 is less or equal to X is less or equal to 57 equals 34% as an example of a probability interval that we know from this activity, but it includes the mean as the minimum. Include at least 7 intervals and their probabilities.

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Answer to a math question Communication and Thinking The time it takes to drive from Orangeville to the Vaughan Mills Mall is normally distributed with a mean of 52 minutes and standard deviation of 5 minutes. What intervals could you estimate, using the knowledge from this activity, that do not include the mean as a max or min? For example, P of X is less than or equal to 52 is 50% is an interval we could estimate. However, it includes the mean as the maximum of the interval. Another example is P of 52 is less or equal to X is less or equal to 57 equals 34% as an example of a probability interval that we know from this activity, but it includes the mean as the minimum. Include at least 7 intervals and their probabilities.

Expert avatar
Dexter
4.7
101 Answers
\text{Mean: } \mu = 52
\text{Standard Deviation: } \sigma = 5

1. Calculate P(X \leq 47)
P(X \leq 47) = P\left(Z \leq \frac{47-52}{5}\right) = P(Z \leq -1)
P(Z \leq -1) = 0.1587
P(X \leq 47) = 0.1587

2. Calculate P(X \leq 42)
P(X \leq 42) = P\left(Z \leq \frac{42-52}{5}\right) = P(Z \leq -2)
P(Z \leq -2) = 0.0228
P(X \leq 42) = 0.0228

3. Calculate P(X \geq 57)
P(X \geq 57) = P\left(Z \geq \frac{57-52}{5}\right) = P(Z \geq 1)
P(Z \geq 1) = 0.1587
P(X \geq 57) = 0.1587

4. Calculate P(X \geq 62)
P(X \geq 62) = P\left(Z \geq \frac{62-52}{5}\right) = P(Z \geq 2)
P(Z \geq 2) = 0.0228
P(X \geq 62) = 0.0228

5. Calculate P(42 \leq X \leq 47)
P(42 \leq X \leq 47) = P\left(\frac{42-52}{5} \leq Z \leq \frac{47-52}{5}\right) = P(-2 \leq Z \leq -1)
P(-2 \leq Z \leq -1) = 0.8413 - 0.1587 = 0.1359
P(42 \leq X \leq 47) = 0.1359

6. Calculate P(57 \leq X \leq 62)
P(57 \leq X \leq 62) = P\left(\frac{57-52}{5} \leq Z \leq \frac{62-52}{5}\right) = P(1 \leq Z \leq 2)
P(1 \leq Z \leq 2) = 0.8413 - 0.1587 = 0.1359
P(57 \leq X \leq 62) = 0.1359

7. Calculate P(47 \leq X \leq 57)
P(47 \leq X \leq 57) = P\left(\frac{47-52}{5} \leq Z \leq \frac{57-52}{5}\right) = P(-1 \leq Z \leq 1)
P(-1 \leq Z \leq 1) = 0.8413 - 0.1587 = 0.6827
P(47 \leq X \leq 57) = 0.6827

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