Question

question
when avani commutes to work, the amount of time it takes her to arrive is normally distributed with a mean of 32 minutes and a standard deviation of 3.5 minutes. what percentage of her commutes will be between 31 and 34 minutes, to the nearest tenth?
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Explanation:

Step1: Identify the distribution and parameters

The time taken to commute is normally distributed with mean $\mu = 32$ minutes and standard deviation $\sigma = 3.5$ minutes. We need to find $P(31 < X < 34)$, where $X$ is the commute time.

Step2: Calculate the z - scores

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$.

For $x = 31$:
$z_1=\frac{31 - 32}{3.5}=\frac{- 1}{3.5}\approx - 0.29$

For $x = 34$:
$z_2=\frac{34 - 32}{3.5}=\frac{2}{3.5}\approx0.57$

Step3: Find the probabilities corresponding to the z - scores

We use the standard normal distribution table (or a calculator with normal distribution functions).

The probability that $Z < z_2$ (where $Z$ is the standard normal variable) for $z_2 = 0.57$ is approximately $P(Z < 0.57)=0.7157$

The probability that $Z < z_1$ (where $z_1=- 0.29$) is approximately $P(Z < - 0.29)=0.3859$

Step4: Calculate the probability between the two z - scores

To find $P(31 < X < 34)$, we calculate $P(z_1 < Z < z_2)=P(Z < z_2)-P(Z < z_1)$

$P(31 < X < 34)=0.7157 - 0.3859 = 0.3298$

Step5: Convert the probability to a percentage

To convert the probability to a percentage, we multiply by 100: $0.3298\times100 = 32.98\%\approx33.0\%$

Answer:

The percentage of her commutes between 31 and 34 minutes is approximately $\boxed{33.0\%}$