Solved on Mar 18, 2024

Find the 28th percentile, middle 96% range, and interquartile range for the number of chocolate chips in a bag, where the number is approximately normally distributed with $\mu=1262$ and $\sigma=118$ .

STEP 1

Assumptions
1. The distribution of chocolate chips in a bag is approximately normal.
2. The mean number of chocolate chips is 1262.
3. The standard deviation is 118.
4. For part (a), we need to find the 28th percentile of this normal distribution.
5. For part (b), we need to find the chocolate chip count that corresponds to the middle 96% of bags.
6. For part (c), we need to calculate the interquartile range (IQR), which is the difference between the 75th percentile (Q3) and the 25th percentile (Q1).

STEP 2

For part (a), we need to find the z-score that corresponds to the 28th percentile in the standard normal distribution.

STEP 3

Using the standard normal distribution table, we look up the z-score that corresponds to the area to the left (cumulative probability) closest to 0.28.

STEP 4

The z-score for the 28th percentile is approximately -0.58. This means that the 28th percentile is 0.58 standard deviations below the mean.

STEP 5

Now we convert the z-score to the actual number of chocolate chips using the formula:
$X = \mu + z \cdot \sigma$
where $X$ is the number of chocolate chips, $\mu$ is the mean, $z$ is the z-score, and $\sigma$ is the standard deviation.

STEP 6

Plug in the values for $\mu$ , $z$ , and $\sigma$ to find the 28th percentile in terms of chocolate chips.
$X = 1262 + (-0.58) \cdot 118$

STEP 7

Calculate the value of $X$ .
$X = 1262 - 0.58 \cdot 118$
$X = 1262 - 68.44$
$X \approx 1193.56$

STEP 8

For part (b), we need to find the chocolate chip counts that correspond to the 2nd percentile and the 98th percentile, since these will mark the boundaries of the middle 96% of bags.

STEP 9

First, we find the z-score for the 2nd percentile using the standard normal distribution table.

STEP 10

The z-score for the 2nd percentile is approximately -2.05.

STEP 11

Now we find the z-score for the 98th percentile. Since the standard normal distribution is symmetric, the z-score for the 98th percentile is the negative of the z-score for the 2nd percentile.
$z_{98th} = -z_{2nd}$

STEP 12

Thus, the z-score for the 98th percentile is approximately 2.05.

STEP 13

We now convert these z-scores to actual chocolate chip counts using the formula from STEP_5.

STEP 14

Calculate the chocolate chip count for the 2nd percentile.
$X_{2nd} = 1262 + (-2.05) \cdot 118$

STEP 15

Calculate the chocolate chip count for the 98th percentile.
$X_{98th} = 1262 + (2.05) \cdot 118$

STEP 16

Calculate the values of $X_{2nd}$ and $X_{98th}$ .
$X_{2nd} = 1262 - 2.05 \cdot 118$
$X_{2nd} = 1262 - 241.9$
$X_{2nd} \approx 1020.1$

STEP 17

Calculate the value of $X_{98th}$ .
$X_{98th} = 1262 + 2.05 \cdot 118$
$X_{98th} = 1262 + 241.9$
$X_{98th} \approx 1503.9$

STEP 18

For part (c), we need to calculate the interquartile range (IQR), which is the difference between the 75th percentile (Q3) and the 25th percentile (Q1).

STEP 19

Find the z-score for the 25th percentile (Q1) using the standard normal distribution table.

STEP 20

The z-score for the 25th percentile is approximately -0.674.

STEP 21

Find the z-score for the 75th percentile (Q3). Since the distribution is symmetric, the z-score for the 75th percentile is the negative of the z-score for the 25th percentile.
$z_{75th} = -z_{25th}$

STEP 22

Thus, the z-score for the 75th percentile is approximately 0.674.

STEP 23

Convert these z-scores to actual chocolate chip counts using the formula from STEP_5.

STEP 24

Calculate the chocolate chip count for the 25th percentile (Q1).
$X_{25th} = 1262 + (-0.674) \cdot 118$

STEP 25

Calculate the chocolate chip count for the 75th percentile (Q3).
$X_{75th} = 1262 + (0.674) \cdot 118$

STEP 26

Calculate the values of $X_{25th}$ and $X_{75th}$ .
$X_{25th} = 1262 - 0.674 \cdot 118$
$X_{25th} = 1262 - 79.532$
$X_{25th} \approx 1182.468$

STEP 27

Calculate the value of $X_{75th}$ .
$X_{75th} = 1262 + 0.674 \cdot 118$
$X_{75th} = 1262 + 79.532$
$X_{75th} \approx 1341.532$

STEP 28

Now, calculate the interquartile range (IQR) by subtracting $X_{25th}$ from $X_{75th}$ .
$IQR = X_{75th} - X_{25th}$
$IQR = 1341.532 - 1182.468$
$IQR \approx 159.064$
The 28th percentile for the number of chocolate chips in a bag is approximately 1194 chips. The number of chocolate chips in a bag that make up the middle 96% of bags ranges from approximately 1020 to 1504 chips. The interquartile range of the number of chocolate chips in a bag of chocolate chip cookies is approximately 159 chips.

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Find the 28th percentile, middle 96% range, and interquartile range for the number of chocolate chips in a bag, where the number is approximately normally distributed with μ=1262\mu=1262μ=1262 and σ=118\sigma=118σ=118.

STEP 1

STEP 2

For part (a), we need to find the z-score that corresponds to the 28th percentile in the standard normal distribution.

STEP 3

Using the standard normal distribution table, we look up the z-score that corresponds to the area to the left (cumulative probability) closest to 0.28.

STEP 4

The z-score for the 28th percentile is approximately -0.58. This means that the 28th percentile is 0.58 standard deviations below the mean.

STEP 5

Now we convert the z-score to the actual number of chocolate chips using the formula:X=μ+z⋅σ X = \mu + z \cdot \sigma X=μ+z⋅σwhere X X X is the number of chocolate chips, μ \mu μ is the mean, z z z is the z-score, and σ \sigma σ is the standard deviation.

STEP 6

Plug in the values for μ \mu μ, z z z, and σ \sigma σ to find the 28th percentile in terms of chocolate chips.X=1262+(−0.58)⋅118 X = 1262 + (-0.58) \cdot 118 X=1262+(−0.58)⋅118

STEP 7

Calculate the value of X X X.X=1262−0.58⋅118 X = 1262 - 0.58 \cdot 118 X=1262−0.58⋅118X=1262−68.44 X = 1262 - 68.44 X=1262−68.44X≈1193.56 X \approx 1193.56 X≈1193.56

STEP 8

For part (b), we need to find the chocolate chip counts that correspond to the 2nd percentile and the 98th percentile, since these will mark the boundaries of the middle 96% of bags.

STEP 9

First, we find the z-score for the 2nd percentile using the standard normal distribution table.

STEP 10

The z-score for the 2nd percentile is approximately -2.05.

STEP 11

Now we find the z-score for the 98th percentile. Since the standard normal distribution is symmetric, the z-score for the 98th percentile is the negative of the z-score for the 2nd percentile.z98th=−z2nd z_{98th} = -z_{2nd} z98th​=−z2nd​

STEP 12

Thus, the z-score for the 98th percentile is approximately 2.05.

STEP 13

We now convert these z-scores to actual chocolate chip counts using the formula from STEP_5.

STEP 14

Calculate the chocolate chip count for the 2nd percentile.X2nd=1262+(−2.05)⋅118 X_{2nd} = 1262 + (-2.05) \cdot 118 X2nd​=1262+(−2.05)⋅118

STEP 15

Calculate the chocolate chip count for the 98th percentile.X98th=1262+(2.05)⋅118 X_{98th} = 1262 + (2.05) \cdot 118 X98th​=1262+(2.05)⋅118

STEP 16

Calculate the values of X2nd X_{2nd} X2nd​ and X98th X_{98th} X98th​.X2nd=1262−2.05⋅118 X_{2nd} = 1262 - 2.05 \cdot 118 X2nd​=1262−2.05⋅118X2nd=1262−241.9 X_{2nd} = 1262 - 241.9 X2nd​=1262−241.9X2nd≈1020.1 X_{2nd} \approx 1020.1 X2nd​≈1020.1

STEP 17

Calculate the value of X98th X_{98th} X98th​.X98th=1262+2.05⋅118 X_{98th} = 1262 + 2.05 \cdot 118 X98th​=1262+2.05⋅118X98th=1262+241.9 X_{98th} = 1262 + 241.9 X98th​=1262+241.9X98th≈1503.9 X_{98th} \approx 1503.9 X98th​≈1503.9

STEP 18

For part (c), we need to calculate the interquartile range (IQR), which is the difference between the 75th percentile (Q3) and the 25th percentile (Q1).

STEP 19

Find the z-score for the 25th percentile (Q1) using the standard normal distribution table.

STEP 20

The z-score for the 25th percentile is approximately -0.674.

STEP 21

Find the z-score for the 75th percentile (Q3). Since the distribution is symmetric, the z-score for the 75th percentile is the negative of the z-score for the 25th percentile.z75th=−z25th z_{75th} = -z_{25th} z75th​=−z25th​

STEP 22

Thus, the z-score for the 75th percentile is approximately 0.674.

STEP 23

Convert these z-scores to actual chocolate chip counts using the formula from STEP_5.

STEP 24

Calculate the chocolate chip count for the 25th percentile (Q1).X25th=1262+(−0.674)⋅118 X_{25th} = 1262 + (-0.674) \cdot 118 X25th​=1262+(−0.674)⋅118

STEP 25

Calculate the chocolate chip count for the 75th percentile (Q3).X75th=1262+(0.674)⋅118 X_{75th} = 1262 + (0.674) \cdot 118 X75th​=1262+(0.674)⋅118

STEP 26

Calculate the values of X25th X_{25th} X25th​ and X75th X_{75th} X75th​.X25th=1262−0.674⋅118 X_{25th} = 1262 - 0.674 \cdot 118 X25th​=1262−0.674⋅118X25th=1262−79.532 X_{25th} = 1262 - 79.532 X25th​=1262−79.532X25th≈1182.468 X_{25th} \approx 1182.468 X25th​≈1182.468

STEP 27

Calculate the value of X75th X_{75th} X75th​.X75th=1262+0.674⋅118 X_{75th} = 1262 + 0.674 \cdot 118 X75th​=1262+0.674⋅118X75th=1262+79.532 X_{75th} = 1262 + 79.532 X75th​=1262+79.532X75th≈1341.532 X_{75th} \approx 1341.532 X75th​≈1341.532

STEP 28

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Find the 28th percentile, middle 96% range, and interquartile range for the number of chocolate chips in a bag, where the number is approximately normally distributed with $\mu=1262$ and $\sigma=118$ .

Now we convert the z-score to the actual number of chocolate chips using the formula:
$X = \mu + z \cdot \sigma$
where $X$ is the number of chocolate chips, $\mu$ is the mean, $z$ is the z-score, and $\sigma$ is the standard deviation.

Plug in the values for $\mu$ , $z$ , and $\sigma$ to find the 28th percentile in terms of chocolate chips.
$X = 1262 + (-0.58) \cdot 118$

Calculate the value of $X$ .
$X = 1262 - 0.58 \cdot 118$
$X = 1262 - 68.44$
$X \approx 1193.56$

Now we find the z-score for the 98th percentile. Since the standard normal distribution is symmetric, the z-score for the 98th percentile is the negative of the z-score for the 2nd percentile.
$z_{98th} = -z_{2nd}$

Calculate the chocolate chip count for the 2nd percentile.
$X_{2nd} = 1262 + (-2.05) \cdot 118$

Calculate the chocolate chip count for the 98th percentile.
$X_{98th} = 1262 + (2.05) \cdot 118$

Calculate the values of $X_{2nd}$ and $X_{98th}$ .
$X_{2nd} = 1262 - 2.05 \cdot 118$
$X_{2nd} = 1262 - 241.9$
$X_{2nd} \approx 1020.1$

Calculate the value of $X_{98th}$ .
$X_{98th} = 1262 + 2.05 \cdot 118$
$X_{98th} = 1262 + 241.9$
$X_{98th} \approx 1503.9$

Find the z-score for the 75th percentile (Q3). Since the distribution is symmetric, the z-score for the 75th percentile is the negative of the z-score for the 25th percentile.
$z_{75th} = -z_{25th}$

Calculate the chocolate chip count for the 25th percentile (Q1).
$X_{25th} = 1262 + (-0.674) \cdot 118$

Calculate the chocolate chip count for the 75th percentile (Q3).
$X_{75th} = 1262 + (0.674) \cdot 118$

Calculate the values of $X_{25th}$ and $X_{75th}$ .
$X_{25th} = 1262 - 0.674 \cdot 118$
$X_{25th} = 1262 - 79.532$
$X_{25th} \approx 1182.468$

Calculate the value of $X_{75th}$ .
$X_{75th} = 1262 + 0.674 \cdot 118$
$X_{75th} = 1262 + 79.532$
$X_{75th} \approx 1341.532$