Answer:
The second choice: Approximately [tex]65.2\%[/tex] of the pretzel bags here will contain between 225 and 245 pretzels.
Step-by-step explanation:
This explanation uses a z-score table where each [tex]z[/tex] entry has two decimal places.
Let [tex]\mu[/tex] represent the mean of a normal distribution of variable [tex]X[/tex]. Let [tex]\sigma[/tex] be the standard deviation of the distribution. The z-score for the observation [tex]x[/tex] would be:
[tex]\displaystyle z = \frac{x - \mu}{\sigma}[/tex].
In this question,
Calculate the z-score for [tex]x_1 = 225[/tex] and [tex]x_2 = 245[/tex]. Keep in mind that each entry in the z-score table here has two decimal places. Hence, round the results below so that each contains at least two decimal places.
[tex]\begin{aligned} z_1 &= \frac{x_1 - \mu}{\sigma} \\ &= \frac{225 - 240}{9.3} \approx -1.61\end{aligned}[/tex].
[tex]\begin{aligned} z_2 &= \frac{x_2 - \mu}{\sigma} \\ &= \frac{245 - 240}{9.3} \approx 0.54\end{aligned}[/tex].
The question is asking for the probability [tex]P(225 \le X \le 245)[/tex] (where [tex]X[/tex] is between two values.) In this case, that's the same as [tex]P(-1.61 \le Z \le 0.54)[/tex].
Keep in mind that the probabilities on many z-table correspond to probability of [tex]P(Z \le z)[/tex] (where [tex]Z[/tex] is no greater than one value.) Therefore, apply the identity [tex]P(z_1 \le Z \le z_2) = P(Z \le z_2) - P(Z \le z_1)[/tex] to rewrite [tex]P(-1.61 \le Z \le 0.54)[/tex] as the difference between two probabilities:
[tex]P(-1.61 \le Z \le 0.54) = P(Z \le 0.54) - P(Z \le -1.61)[/tex].
Look up the z-table for [tex]P(Z \le 0.54)[/tex] and [tex]P(Z \le -1.61)[/tex]:
[tex]\begin{aligned}& P(225 \le X \le 245) \\ &= P\left(\frac{225 - 240}{9.3} \le Z \le \frac{245 - 240}{9.3}\right)\\&\approx P(-1.61 \le Z \le 0.54) \\ &= P(Z \le 0.54) - P(Z \le -1.61)\\ &\approx 0.70540 - 0.05370 \\& \approx 0.65.2 \\ &= 65.2\% \end{aligned}[/tex].
