Question 1179726: The weight distribution of a sample of adults with physical inabilities is approximately normal, with mean 72 and standard deviation 8. Find an interval of values around the mean such that it
Includes 95 percent of the observed values
Includes almost all observed values (and thus coincides with the range, min-max).
Includes 50 percent of the observed values.
Snakes deposit chemical trails as they travel through their habitats. These trails are often detected and recognized by lizards, which are potential prey. The ability to recognize their predators via tongue flicks can often mean life or death for lizards. Scientist from Kenyatta University were interested in quantifying the responses of Juveniles of the common lizard to the natural predator cues to determine whether the behavior is learned or congenital. Seventeen juvenile common lizards were exposed to the chemical cues of the viper snake. Their responses, in number of tongue flicks per 20 minutes, are presented in the following table:
425 510 629 236 654 200 710 276 633
501 811 332 424 674 676 662 694
Find a 90% confidence interval for the mean number of tongue flicks per 20 minutes for all juvenile common lizards. Assume a population standard deviation 190.
Interpret your answer from part (i).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
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