Question

6.49 Delaying Adulthood. In the paper, "Delayed Metamorphosis of a Tropical Reef Fish (Acanthurus triostegus): A Field Experiment" (Marine Ecology Progress Series, Vol. 176, pp. 25-38). M. McCormick studied larval duration of the convict surgeonfish. a common tropical reef fish. This fish has been found to delay metamorphosis into adulthood by extending its larval phase, a delay that often leads to enhanced survivorship in the species by increasing the chances of finding suitable habitat. Duration of the larval phase for convict surgeonfish is normally distributed with mean 53 days and standard deviation 3.4 days. Let $x$denote larval-phase duration for convict surgeonfish.

a. Sketch the normal curve for the variable $x$.
b. Simulate 1500 observations of $x$. (Note; Users of the T7-83/84 Plus should simulate 750 observations.)
c. Approximately what values would you expect for the sample mean and sample standard deviation of the 1500 observations? Explain your answers.
d. Obtain the sample mean and sample standard deviation of the 1500 observations, and compare your answers to your estimates in part (c).
e. Roughly what would you expect a histogram of the 1500 observations to look like? Explain your answer.
f. Obtain a histogram of the 1500 observations, and compare your result to your expectation in part (e).

Short answer

(a) The $x$variable's normal curve as follows:

(b) The following is an example of a possible outcome:

(c) For $x$, the mean of the above $1500$observations is $52.96379$, and the standard deviation is $3.451587$, which is near to $53$and $3.4$.

(d) The sample mean of the $1500$observations is $53.1727$, and the standard deviation is $3.4223$.
(e) The histogram of the given observation resembles that of a population curve.

(f) The histogram of 1500 observations as follows:

Step-by-step solution

Part (a) Step 1: Given information

To sketch the normal curve for the variable $x$.

Part (a) Step 2: Explanation

The length of a convict surgeonfish's larval period is regularly distributed, with a mean of $53$days and a standard deviation of $3.4$days. It's a representation of a normal distribution in graphical form. If $x$is a continuous random variable with a mean of $\mu$and a standard deviation of $\sigma$, then a normal curve with random variable $x$has the following equation:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0$

Furthermore, a normal curve with random variable $z$has the following equation:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

where $z$has a mean of 0 and a standard deviation of 1 accordingly.

A normal curve typically includes two population parameters: population mean $\mu$and population standard deviation $\sigma$.

$\mu =53$

Therefore, the $x$variable's normal curve as follows:

Part (b) Step 1: Given information

To simulate $1500$observations of $x$. Note that: Users of the $T7-83/84$ Plus should simulate $750$ observations.

Part (b) Step 2: Explanation

Generate $1500$for $x$at random from a normal distribution with a mean of $\mu =53$and a standard deviation of $\mu =3.4$. Use the $NORMINV$command in EXCEL to produce such numbers at random $(RANDO,53,3.4)$. Where $RAND$generates a random number between 0 and 1, and $NORMINV$estimates the number of normal distributions with a probability of a random number between $0$and $1$.
The following is an example of a possible outcome:

Part (c) Step 1: Given information

To explain the values approximately that expect for the sample mean and sample standard deviation of the $1500$ observations.

Part (c) Step 2: Explanation

Calculate the mean and standard deviation of the above $1500$observations of $x$ using EXCEL.
The following is an example of a possible outcome $b$:

Mean	52.96379
Standard Error	0.089149
Median	52.96294
Mode	58.89265
Standard Deviation	3.451587
Sample Variance	11.91345
Kurtosis	-0.11567
Skewness	0.022045
Range	24.12871
Minimum	40.2786
Maximum	64.4073
Sum	79392.73
Count	1499

As a result, for $x$, the mean of the above $1500$ observations is $52.96379$, and the standard deviation is $3.451587$, which is near to $53$ and $3.4$.

Part (d) Step 1: Given information

To obtain the sample mean and sample standard deviation of the $1500$ observations, and compare the answers which estimates in part (c).

Part (d) Step 2: Explanation

Since, the number of observations is $n$ as follows:
$n=1500$
Because the mean is equal to the entire number of observations divided by the total number of observations,
$\overline{x}=\frac{{\sum }^x}{n}$

$\approx 53.1727$

The variance is calculated by dividing the sum of squared deviations from the mean by $n-1$.

Therefore, the standard deviation is equal to the square roots of variance.

$s=\sqrt{\frac{(53.1090-53.1727{)}^2+(52.6599-53.1727{)}^2+\dots }{1500-1}}\approx 3.4223$

As a result, the sample mean of the $1500$observations is $53.1727$, and the standard deviation is $3.4223$.

Part (e) Given information

To explain that roughly that expect a histogram of the $1500$ observations.

Step (e) Part 2: Explanation

Let , the histogram of 1500 observations resembles that of a population curve.

Asa result, the histogram of the given observation resembles that of a population curve.

Part (f) Step 1: Given information

To obtain a histogram of the $1500$ observations and to compare the result of expectation in part (e).

Part (f) Step 2: Explanation

The histogram of 1500 observations as follows:

As a result, the histogram is bell-shaped, as can be observed (unimodal and symmetric).