# What is the probability that a deck of $52$ cards is more than $0.55$ inches in thickness?

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?

• 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 – Satish Ramanathan Jan 25 at 14:00
• You just forgot a zero in your computation : $0,00052$ instead of $0,000052$. – Ayoub Jan 25 at 14:03

There is something wrong with your $$z$$-score:
• $$X =\sum_{k=1}^{52}Y_k$$ with $$Y_k \sim N(0.01,52\cdot 10^{-6})$$
• $$\Rightarrow X \sim N(0.52,52\cdot52\cdot 10^{-6})$$ $$\Rightarrow P(X>0.55) = P\left( \frac{X-0.52}{52\cdot 10^{-3}} > \frac{0.55-0.52}{52\cdot 10^{-3}}\right) \approx P\left( Z > 0.5769\right) \mbox{ where } Z \sim N(0,1)$$