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.

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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.

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84 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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