Question 1179726
Let's solve both parts of this problem:

**Part 1: Weight Distribution of Adults with Physical Inabilities**

* Mean (μ) = 72
* Standard deviation (σ) = 8
* Distribution: Normal

**(a) Includes 95 percent of the observed values:**

* For a 95% confidence interval in a normal distribution, we use a z-score of approximately 1.96.
* Margin of error (E) = z * σ = 1.96 * 8 = 15.68
* Interval: μ ± E = 72 ± 15.68
* Lower bound: 72 - 15.68 = 56.32
* Upper bound: 72 + 15.68 = 87.68
* Interval: (56.32, 87.68)

**(b) Includes almost all observed values (min-max):**

* "Almost all" typically refers to about 99.7% of the data, which corresponds to 3 standard deviations from the mean.
* Margin of error (E) = 3 * σ = 3 * 8 = 24
* Interval: μ ± E = 72 ± 24
* Lower bound: 72 - 24 = 48
* Upper bound: 72 + 24 = 96
* Interval: (48, 96)

**(c) Includes 50 percent of the observed values:**

* For 50% of the data, we're looking for the interquartile range (IQR).
* The IQR is the range between the 25th and 75th percentiles.
* For a normal distribution, the z-scores corresponding to the 25th and 75th percentiles are approximately ±0.6745.
* Margin of error (E) = 0.6745 * 8 ≈ 5.396
* Interval: 72 ± 5.396
* Lower bound: 72 - 5.396 = 66.604
* Upper bound: 72 + 5.396 = 77.396
* Interval: (66.604, 77.396)

**Part 2: Tongue Flicks of Juvenile Lizards**

* Data: 425, 510, 629, 236, 654, 200, 710, 276, 633, 501, 811, 332, 424, 674, 676, 662, 694
* Sample size (n) = 17
* Population standard deviation (σ) = 190
* Confidence level = 90%

**(i) Find a 90% confidence interval:**

1.  **Calculate the Sample Mean (x̄):**
    * x̄ = (425 + 510 + 629 + 236 + 654 + 200 + 710 + 276 + 633 + 501 + 811 + 332 + 424 + 674 + 676 + 662 + 694) / 17
    * x̄ = 8741 / 17 ≈ 514.18

2.  **Find the Critical Z-Value:**
    * For a 90% confidence level, α = 1 - 0.90 = 0.10.
    * α/2 = 0.05.
    * The z-value corresponding to 0.95 (1 - 0.05) is approximately 1.645.

3.  **Calculate the Margin of Error (E):**
    * E = z * (σ / √n)
    * E = 1.645 * (190 / √17)
    * E ≈ 1.645 * (190 / 4.123)
    * E ≈ 75.71

4.  **Construct the Confidence Interval:**
    * Lower Bound = x̄ - E = 514.18 - 75.71 ≈ 438.47
    * Upper Bound = x̄ + E = 514.18 + 75.71 ≈ 589.89
    * Confidence Interval: (438.47, 589.89)

**(ii) Interpret the answer:**

* We are 90% confident that the true mean number of tongue flicks per 20 minutes for all juvenile common lizards exposed to viper snake chemical cues is between 438.47 and 589.89.
* This means that if we were to repeat this experiment many times and calculate a 90% confidence interval each time, approximately 90% of those intervals would contain the true population mean number of tongue flicks.