# Chapter 6 - Normal Probability Distributions - 6-7 Normal as Approximation to Binomial - Beyond the Basics: 26

241 reservations.

#### Work Step by Step

By using the table, the z-score corresponding to 0.5: z=1.645. $mean=n\cdot p=213\cdot 0.0995=21.1935.$ $standard \ deviation=\sqrt{n\cdot p \cdot q}=\sqrt{n\cdot p \cdot (1-p)}=\sqrt{213\cdot 0.0995 \cdot 0.9005}=4.3686.$ Hence the corresponding value:$mean+z⋅standard \ deviation=21.1935+1.645⋅4.3686\approx28.$ Hence we can accept 213+28=241 reservations.

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