A professor has three errands to take care of in the...

A professor has three errands to take care of in the Administration Building. Let $X_{i}=$ the time that it takes for the $i$ th errand $(i=1,2,3),$ and let $X_{4}=$ the in minutes that spends walking to and from the building and between each errand. Suppose the $X_{i}$ s are independent, normally distributed, with the following means and standard deviations: $\mu_{1}=15, \sigma_{1}=4$ $\mu_{2}=5, \sigma_{2}=1, \mu_{3}=8, \sigma_{3}=2, \mu_{4}=12, \sigma_{4}=3 .$ She plans to leave her office at precisely $10 : 00$ a.m. and wishes to post a note on her door that reads, "I will return by $t$ a.m." What time $t$ should she write down if she wants the probability of her arriving after $t$ to be .01$?$

A professor has three errands to take care of in the Administration Building. Let $X_{i}=$ the time
that it takes for the $i$ th errand $(i=1,2,3),$ and let $X_{4}=$ the in minutes that spends walking to and from the building and between each errand. Suppose the $X_{i}$ s are independent, normally distributed, with the following means and standard deviations: $\mu_{1}=15, \sigma_{1}=4$ $\mu_{2}=5, \sigma_{2}=1, \mu_{3}=8, \sigma_{3}=2, \mu_{4}=12, \sigma_{4}=3 .$ She plans to leave her office at precisely $10 : 00$ a.m. and wishes to post a note on her door that reads, "I will return by $t$ a.m." What time $t$ should she write down if she wants the probability of her arriving after $t$ to be .01$?$

Key Concepts

Normal Distribution

The normal distribution is a continuous probability distribution characterized by its symmetric bell shape, where the mean defines the center and the standard deviation determines the spread of the data. It is widely used in many fields to model real-world phenomena that cluster around an average value, making it pivotal in statistical analysis and probability theory.

Sum of Independent Normal Variables

When independent random variables are normally distributed, their sum is also normally distributed. The resulting distribution has a mean equal to the sum of the individual means and a variance equal to the sum of the individual variances. This concept is crucial in practical applications, such as aggregating time durations or error measurements, to determine the overall distribution of a combined process.

Standardization of a Normal Variable

Standardization is the process of converting a normal random variable to a standard normal variable (with mean zero and standard deviation one) by subtracting the mean and dividing by the standard deviation. This transformation allows one to use standardized tables or computational tools to easily calculate probabilities and quantiles associated with any normal distribution.

Quantiles and Inverse Cumulative Distribution Function

Quantiles determine the value below which a certain percentage of observations in a distribution fall. The inverse cumulative distribution function (or quantile function) is used to find these values. In contexts such as setting deadlines or risk assessment, quantile calculations provide threshold estimates (for example, the 99th percentile) that ensure only a small 'tail' probability (e.g., 1%) exceeds this critical value.

Transcript

00:01 In this problem, we have a professor who's going out to run four errands, which we know the mean and standard deviation of, and wants to know how long it would take her to run the errands and return to the office, with the probability of the time derives to be 0 .01 or in the 99th percentile.

00:17 So we can go ahead and start by creating a variable y, which is equal to the total time it would take to run the errands and return back to the office.

00:26 Now, in this problem, we also have a variable x subi, which is, equal to the time it took to run each it's errand.

00:34 So then we can just say that y would then equal to sum from i equals 1 to i equals 4 of x sub i now because each x sub i variable is both independent and normally distributed we can then say that y is then normally distributed, which that implies that we can go ahead and calculate both the mean and standard deviation of y because y is normally distributed to calculate the mean of y, we can just say that the mean of y is equal to the sum from i equals 1 to 4 of the mean of x sub i.

01:17 So that then gives us um, mean of 1 plus the mean of 2 plus the mean of 3 plus the mean of 4 has stated in the problem, which those values are 15 plus 5.

01:33 Plus 8 plus 12 and throwing this into our calculator we get 40 as our mean of y.

01:43 Unfortunately, the standard deviation of y is a little harder to calculate than the mean of y was.

01:50 To calculate the standard deviation of y, we need to first to get the variance of y by summing together the variance of x and then we can take that square root.

02:00 So if we start here, we can say the variance of y is then equal.

02:05 To the sum from i equals one to four of the variance of x sub i now the variance is just equal to the standard deviation squared so we can then just say the standard deviation of one squared plus the standard deviation of two squared plus the standard deviation of three squared plus the standard deviation of four squared now those values are given in the problem so we can just substitute those in as 4 squared plus 1 squared plus 2 squared plus 3 squared to give us a total answer of 30 for the variance of y...

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