Many sharks enter a state of tonic immobility when inverted. Suppose that in a particular species of sharksthe time a shark remains in a state of tonic immobility when inverted is normally distributed with mean 11.2minutes and standard deviation 1.1 minutes.
a. If a biologist induces a state of tonic immobility in such a shark in order to study it, find theprobability that the shark will remain in this state for between 10 and 13 minutes.
b. When a biologist wishes to estimate the mean time that such sharks stay immobile by inducing tonic immobility in each of a sample of 12 sharks, find the probability that mean time ofimmobility in the sample will be between 10 and 13 minutes.
The correct answer and explanation is :
Part (a) — Probability that a shark remains in tonic immobility for between 10 and 13 minutes:
In this part, we are given that the time a shark remains in a state of tonic immobility is normally distributed with a mean of 11.2 minutes and a standard deviation of 1.1 minutes. We are asked to find the probability that the shark remains in this state for between 10 and 13 minutes.
To solve this, we can use the standard normal distribution by first transforming the raw times (10 and 13 minutes) into Z-scores using the formula:
[
Z = \frac{X – \mu}{\sigma}
]
Where:
- ( X ) is the value of interest (10 or 13 minutes),
- ( \mu ) is the mean (11.2 minutes),
- ( \sigma ) is the standard deviation (1.1 minutes).
Step 1: Calculate the Z-scores
For ( X = 10 ) minutes:
[
Z_1 = \frac{10 – 11.2}{1.1} = \frac{-1.2}{1.1} \approx -1.09
]
For ( X = 13 ) minutes:
[
Z_2 = \frac{13 – 11.2}{1.1} = \frac{1.8}{1.1} \approx 1.64
]
Step 2: Find the probabilities corresponding to these Z-scores using the standard normal distribution table.
From the standard normal distribution table:
- The cumulative probability for ( Z_1 = -1.09 ) is approximately 0.1379.
- The cumulative probability for ( Z_2 = 1.64 ) is approximately 0.9495.
Step 3: Calculate the probability between these Z-scores:
The probability that the shark remains in tonic immobility for between 10 and 13 minutes is the difference between the cumulative probabilities:
[
P(10 \leq X \leq 13) = P(Z_2) – P(Z_1) = 0.9495 – 0.1379 = 0.8116
]
So, the probability is approximately 0.8116 or 81.16%.
Part (b) — Probability that the mean time of immobility in a sample of 12 sharks is between 10 and 13 minutes:
In this part, we are considering the mean time of immobility for a sample of 12 sharks. The sampling distribution of the sample mean will follow a normal distribution with:
- Mean ( \mu = 11.2 ) minutes,
- Standard deviation ( \sigma_{\text{mean}} = \frac{\sigma}{\sqrt{n}} = \frac{1.1}{\sqrt{12}} ).
Step 1: Calculate the standard error of the mean:
[
\sigma_{\text{mean}} = \frac{1.1}{\sqrt{12}} \approx \frac{1.1}{3.464} \approx 0.318
]
Step 2: Standardize the values (find the Z-scores):
For ( X = 10 ) minutes:
[
Z_1 = \frac{10 – 11.2}{0.318} = \frac{-1.2}{0.318} \approx -3.77
]
For ( X = 13 ) minutes:
[
Z_2 = \frac{13 – 11.2}{0.318} = \frac{1.8}{0.318} \approx 5.66
]
Step 3: Find the cumulative probabilities for these Z-scores:
From the standard normal distribution table:
- The cumulative probability for ( Z_1 = -3.77 ) is approximately 0.00008 (almost 0).
- The cumulative probability for ( Z_2 = 5.66 ) is approximately 1.0000 (almost 1).
Step 4: Calculate the probability between these Z-scores:
[
P(10 \leq \bar{X} \leq 13) = P(Z_2) – P(Z_1) = 1.0000 – 0.00008 = 0.99992
]
So, the probability is approximately 0.9999 or 99.99%.
Summary of Results:
- Part (a): The probability that a shark remains in tonic immobility for between 10 and 13 minutes is 81.16%.
- Part (b): The probability that the mean time of immobility in a sample of 12 sharks will be between 10 and 13 minutes is 99.99%.
Explanation:
The key difference between parts (a) and (b) lies in the sampling process. In part (a), we are looking at the individual time of immobility for a single shark, which follows a normal distribution with a mean of 11.2 minutes and a standard deviation of 1.1 minutes. In part (b), we are considering the sample mean of 12 sharks, which has a smaller standard deviation (standard error) because we average the times. This results in a much higher probability that the sample mean will fall between 10 and 13 minutes, reflecting the fact that averages tend to be less variable than individual observations.