Normal distribution question: the time required to assemble bookcases is normally distributed...

[tpupble]

The question is as follows: the time required to assemble bookcases is normally distributed with a mean of 48 minutes and a standard deviation of 10 minutes. Miller buys 10 of these bookcases. What is the probability that the total time it takes him to assemble all 8 bookcases is more than 6 hours?

multiply the mean by 8 to get 384 minutes. The average time taken to assemble 8 bookcases is 384 minutes. The standard deviation formula for a sample (as per my curriculum) is σ/√n , with σ being the population standard deviation and n representing the number in the sample.

so, I input this info into my graphics calculator to find the area to the right of 384.
Lower: 360
Upper: 10000
σ: 10/√8
μ: 384

The resulting probability is 1. But the answer is actually 0.9918 which corresponds to letting σ = 10. Why is it that σ/√n is not used in this case as the standard deviation when the 8 bookcases are a sample of 8 from the overall population of bookcases?

 

[blamocur]

The standard deviation formula for a sample (as per my curriculum) is σ/√n

This looks more like the formula for the standard deviation of the average value, not the sum.

 

[blamocur]

Miller buys 10 of these bookcases. What is the probability that the total time it takes him to assemble all 8 bookcases

Miller buys 10 bookcases but assembles only 8?

 

[blamocur]

I get 0.9918 for 9 bookcases using $\sigma\sqrt{n} = 3\sigma$ for standard deviation.

 

[tpupble]

Miller buys 10 bookcases but assembles only 8?

My bad for typo, buys 8 bookcases and assembled 8

 

[tpupble]

I get 0.9918 for 9 bookcases using $\sigma\sqrt{n} = 3\sigma$ for standard

Do you mean the answer is wrong? Could I see your inputs into the technology you used please, am not quite understanding.
As in Lower bound, upper bound, standard deviation and mean on the normal cumulative distribution calculator.

 

[Harry_the_cat]

I disagree with the answer of 0.9918.

If X is the number of minutes to assemble one bookcase then $\displaystyle X \sim N(\mu, \sigma^2)$ where $\displaystyle \mu = 48$ and $\displaystyle \sigma^2 =100$.
In other words, $\displaystyle Exp(X)=48$ and $\displaystyle Var(X)=100$

Using $\displaystyle Exp(aX) = a Exp(X)$ and $\displaystyle Var(aX)=a^2Var(X)$ we get

$\displaystyle Exp(8X) = 8\times48 = 384$ and $\displaystyle Var(8X) = 8^2 \times 100 =6400$ or $\displaystyle sd(8X) = \sqrt{6400} = 80$.

So $\displaystyle 8X \sim N(384, 80^2)$

Now we need to find P(8X>360)

Using Graphics calculator:
Lower = 360
Upper = 1000000
$\displaystyle \sigma$ = 80
$\displaystyle \mu$ = 384

P(8X>360) = 0.6179

I don't believe the question involves taking a sample of size 8.

 

[blamocur]

Var(8X)=8^2×100

Did you mean Var(8X) = 8 x 100 ?

 

[blamocur]

Could I see your inputs into the technology you used

I used Python/SciPy script -- do you have access to those?

 

[blamocur]

I used Python/SciPy script -- do you have access to those?

Even if you don't I am posting the script below in case someone reading this thread wants to double check it. The script computes probabilities for numbers of bookcases in the [6,12] range:

import numpy as np
import scipy.stats

targetTime = 6 * 60;  ## 6 hours
meanTime1  = 48;
stdDev1    = 10;

for numBookCases in range (6,13):
    meanTimeM          = meanTime1 * numBookCases;## Mean time for multiple bookcases:
    relativeTargetTime = targetTime - meanTimeM
    stdDevM            = stdDev1 * np.sqrt(numBookCases);
    normRelTgtTime = (relativeTargetTime) / stdDevM;
    ##................................ Probability of finiishing below target time:
    prob0 = scipy.stats.norm(0,1).cdf (normRelTgtTime);
    ##.................................Probability of finiishing above target time
    prob1 = 1 - prob0;
    print ("%3d %10.8f    (%7.3f)" % (numBookCases, prob1, normRelTgtTime))

 

[BigBeachBanana]

Note that $$8X \neq X_1 + X_2 + ... + X_8$$
Edit: The LHS indicates take 1 random sample and multiply the outcome by 8, whereas the RHS indicates 8 random samples.

Define $S_8 = X_1 + X_2 + ... + X_8$
$$E(S_8) = E(X_1+X_2+...+X_8) = E(X_1) + E(X_2) + ... E(X_8) = 8\times 48 = 384\\ V(S_8) = V(X_1+X_2+...+X_8) = V(X_1) + V(X_2) + ... +V(X_8) =8 \times 100=800\\ \Pr\left(S_8>360\right) = \Pr\left(Z> \dfrac{360-384}{\sqrt{800}}\right) =0.8019$$
$$\Pr\left(S_9>360\right) = \Pr\left(Z> \dfrac{360-9\times48}{\sqrt{900}}\right) =0.9918$$

 

[Harry_the_cat]

Did you mean Var(8X) = 8 x 100 ?

No. $\displaystyle Var(aX)=a^2Var(X)$.

 

[blamocur]

No. $\displaystyle Var(aX)=a^2Var(X)$.

@BigBeachBanana in post #11 stated it better than I, i.e., scaled random variable is different from the sum of independent random variables.

 

[Harry_the_cat]

@BigBeachBanana in post #11 stated it better than I, i.e., scaled random variable is different from the sum of independent random variables.

Yeah ok. I concede.