Question

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%

Answer 1

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}`.