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The thickness of the individual cards produced by a certain playing card manufacturer is normally distributed with mean $0.01$ inches and variance $0.000052$. What is the probability that a deck of $52$ cards is more than $0.55$ inches in thickness? (The thickness of each card is independent of the others).

Solve: \begin{align}P(X>0.55)&=P\left(Z>\frac{\frac{0.55}{52}-0.01}{\frac{\sqrt{0.00052}}{\sqrt{52}}}\right)\\ &=P(Z>0.182)\\ &=0.427\end{align} (from the standard normal tables)

From the book solution it should be $P(Z > 0.58) ≈ 0.28$, but I can't see where I'm wrong, can someone help me?

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    $\begingroup$ I think there is a small mistake, you have to take the square root of (0.000052) instead of (0.00052) and hence the difference $\endgroup$ – Satish Ramanathan Jan 25 at 14:00
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    $\begingroup$ You just forgot a zero in your computation : $0,00052$ instead of $0,000052$. $\endgroup$ – Ayoub Jan 25 at 14:03
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There is something wrong with your $z$-score:

  • $X =\sum_{k=1}^{52}Y_k$ with $Y_k \sim N(0.01,52\cdot 10^{-6})$
  • $\Rightarrow X \sim N(0.52,52\cdot52\cdot 10^{-6})$ $$\Rightarrow P(X>0.55) = P\left( \frac{X-0.52}{52\cdot 10^{-3}} > \frac{0.55-0.52}{52\cdot 10^{-3}}\right) \approx P\left( Z > 0.5769\right) \mbox{ where } Z \sim N(0,1)$$
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I just typed it into my calculator, if what you wrote down is correct then you have forgotten an extra 0 in your answer. Plugging 0.000052 as variance into the formula gives the book answer

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  • $\begingroup$ You are right, thank you! $\endgroup$ – Mark Jacon Jan 25 at 14:02

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