Can someone answer this please.

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

6 0

Answer:

This is the answer I am writing the answer rn

Step-by-step explanation:

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Use technology or a z-score table to answer the question.

Alik [6]

Answer:

The second choice: Approximately $65.2\%$ 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 $z$ entry has two decimal places.

Let $\mu$ represent the mean of a normal distribution of variable $X$ . Let $\sigma$ be the standard deviation of the distribution. The z-score for the observation $x$ would be:

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question,

$\mu = 240$ .
$\sigma = 9.3$ .

Calculate the z-score for $x_1 = 225$ and $x_2 = 245$ . 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.

$\begin{aligned} z_1 &= \frac{x_1 - \mu}{\sigma} \\ &= \frac{225 - 240}{9.3} \approx -1.61\end{aligned}$ .

$\begin{aligned} z_2 &= \frac{x_2 - \mu}{\sigma} \\ &= \frac{245 - 240}{9.3} \approx 0.54\end{aligned}$ .

The question is asking for the probability $P(225 \le X \le 245)$ (where $X$ is between two values.) In this case, that's the same as $P(-1.61 \le Z \le 0.54)$ .

Keep in mind that the probabilities on many z-table correspond to probability of $P(Z \le z)$ (where $Z$ is no greater than one value.) Therefore, apply the identity $P(z_1 \le Z \le z_2) = P(Z \le z_2) - P(Z \le z_1)$ to rewrite $P(-1.61 \le Z \le 0.54)$ as the difference between two probabilities:

$P(-1.61 \le Z \le 0.54) = P(Z \le 0.54) - P(Z \le -1.61)$ .

Look up the z-table for $P(Z \le 0.54)$ and $P(Z \le -1.61)$ :

$P(Z \le 0.54)\approx 0.70540$ .
$P(Z \le -1.61) \approx 0.05370$ .

$\begin{aligned}& P(225 \le X \le 245) \\ &= P\left(\frac{225 - 240}{9.3} \le Z \le \frac{245 - 240}{9.3}\right)\\&\approx P(-1.61 \le Z \le 0.54) \\ &= P(Z \le 0.54) - P(Z \le -1.61)\\ &\approx 0.70540 - 0.05370 \\& \approx 0.65.2 \\ &= 65.2\% \end{aligned}$ .