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

The number of pretzels in a bag is normally distributed with a mean of 240 pretzels and a standard deviation of 9.3 pretzels.

Approximately what percent of the bags of pretzels have between 225 and 245 pretzels?


5.4%

65.2%

70.5%

94.6%

User Max DeLiso
by
4.1k points

1 Answer

5 votes

Answer:

The second choice: Approximately
65.2\% of the pretzel bags here will contain between 225 and 245 pretzels.

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 = (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 &= (x_1 - \mu)/(\sigma) \\ &= (225 - 240)/(9.3) \approx -1.61\end{aligned}.


\begin{aligned} z_2 &= (x_2 - \mu)/(\sigma) \\ &= (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((225 - 240)/(9.3) \le Z \le (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}.

User Qi
by
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