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

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tpupble

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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?
 
I get 0.9918 for 9 bookcases using [imath]\sigma\sqrt{n} = 3\sigma[/imath] for standard deviation.
 
blamocur said:
IMG_1515.jpegI get 0.9918 for 9 bookcases using [imath]\sigma\sqrt{n} = 3\sigma[/imath] for standard
Click to expand...
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.
 
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.
 
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blamocur said:
I used Python/SciPy script -- do you have access to those?
Click to expand...
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:
Code: 
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))
 
Note that [math]8X \neq X_1 + X_2 + ... + X_8[/math]
Edit: The LHS indicates take 1 random sample and multiply the outcome by 8, whereas the RHS indicates 8 random samples.

Define [imath]S_8 = X_1 + X_2 + ... + X_8[/imath]
[math]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 [/math]
[math]\Pr\left(S_9>360\right) = \Pr\left(Z> \dfrac{360-9\times48}{\sqrt{900}}\right) =0.9918[/math]
 
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