Find the measures of variability and any outliers for each set of data. 5. the number of minutes spent reading each night: $$ 31,33,32,34,35,33 $$ 6. the number of fish in each fish tank: $$ 6,5,7,8,5,6,7 $$, $$ 9,8,6 $$

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Alright, I need to find the measures of variability and identify any outliers for the given datasets. Let's tackle them one by one.

**Dataset 1: Number of minutes spent reading each night**
The data points are: 31, 33, 32, 34, 35, 33.

First, I'll calculate the mean to understand the central tendency.
Mean = (31 + 33 + 32 + 34 + 35 + 33) / 6 = 198 / 6 = 33 minutes.

Next, I'll find the variance. Variance measures how much the numbers are spread out from the mean.
Variance = Σ (x_i - mean)^2 / (n - 1)
Calculating each term:
(31 - 33)^2 = 4
(33 - 33)^2 = 0
(32 - 33)^2 = 1
(34 - 33)^2 = 1
(35 - 33)^2 = 4
(33 - 33)^2 = 0
Sum = 4 + 0 + 1 + 1 + 4 + 0 = 10
Variance = 10 / (6 - 1) = 10 / 5 = 2

Standard deviation is the square root of variance.
Standard Deviation = √2 ≈ 1.414 minutes.

Now, to identify outliers, I'll use the IQR method. First, find the first quartile (Q1) and third quartile (Q3).
The data sorted: 31, 32, 33, 33, 34, 35
Q1 = median of first half: (31 + 32) / 2 = 31.5
Q3 = median of second half: (33 + 34) / 2 = 33.5
IQR = Q3 - Q1 = 33.5 - 31.5 = 2

Any data point below Q1 - 1.5 * IQR or above Q3 + 1.5 * IQR is considered an outlier.
Lower bound = 31.5 - 1.5 * 2 = 31.5 - 3 = 28.5
Upper bound = 33.5 + 1.5 * 2 = 33.5 + 3 = 36.5

Looking at the data:
- 31 is below 28.5, so it's an outlier.
- 35 is above 36.5, so it's also an outlier.

**Dataset 2: Number of fish in each fish tank**
The data points are: 6, 5, 7, 8, 5, 6, 7, 9, 8, 6.

First, calculate the mean.
Mean = (6 + 5 + 7 + 8 + 5 + 6 + 7 + 9 + 8 + 6) / 10 = 65 / 10 = 6.5 fish.

Next, find the variance.
Variance = Σ (x_i - mean)^2 / (n - 1)
Calculating each term:
(6 - 6.5)^2 = 0.25
(5 - 6.5)^2 = 2.25
(7 - 6.5)^2 = 0.25
(8 - 6.5)^2 = 2.25
(5 - 6.5)^2 = 2.25
(6 - 6.5)^2 = 0.25
(7 - 6.5)^2 = 0.25
(9 - 6.5)^2 = 6.25
(8 - 6.5)^2 = 2.25
(6 - 6.5)^2 = 0.25
Sum = 0.25 + 2.25 + 0.25 + 2.25 + 2.25 + 0.25 + 0.25 + 6.25 + 2.25 + 0.25 = 16.5
Variance = 16.5 / (10 - 1) = 16.5 / 9 ≈ 1.833

Standard deviation = √1.833 ≈ 1.353 fish.

Now, identify outliers using the IQR method.
Sorted data: 5, 5, 6, 6, 6, 7, 7, 8, 8, 9
Q1 = median of first half: (6 + 6) / 2 = 6
Q3 = median of second half: (8 + 8) / 2 = 8
IQR = 8 - 6 = 2

Lower bound = 6 - 1.5 * 2 = 6 - 3 = 3
Upper bound = 8 + 1.5 * 2 = 8 + 3 = 11

Looking at the data:
- All data points are between 3 and 11, so there are no outliers.

**Summary:**
1. **Minutes spent reading each night:**
   - Mean: 33 minutes
   - Variance: 2
   - Standard Deviation: ≈1.414 minutes
   - Outliers: 31 and 35 minutes

2. **Number of fish in each fish tank:**
   - Mean: 6.5 fish
   - Variance: ≈1.833
   - Standard Deviation: ≈1.353 fish
   - No outliers

**Dataset 1: Number of minutes spent reading each night** - **Mean:** 33 minutes - **Variance:** 2 - **Standard Deviation:** ≈1.414 minutes - **Outliers:** 31 and 35 minutes **Dataset 2: Number of fish in each fish tank** - **Mean:** 6.5 fish - **Variance:** ≈1.833 - **Standard Deviation:** ≈1.353 fish - **Outliers:** None

Find the measures of variability and any outliers for each set of data. 5. the number of minutes spent reading each night: \( 31,33,32,34,35,33 \) 6. the number of fish in each fish tank: \( 6,5,7,8,5,6,7 \), \( 9,8,6 \)

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