Q. 25 - Elementary Statistics

Question

25. Forearm Length. In 1903 , K. Pearson and A. Lee published a paper entitled "On the Laws of Inheritance in Man. I. Inheritance of Physical Characters" (Biometrika, Vol. 2, pp. 357-462). From information presented in that paper, forearm length of men, measured from the elbow to the middle fingertip, is (roughly) normally distributed with a mean of 18.8 inches and a standard deviation of 1.1 inches. Let $x$ denote forearm length, in inches, for men.
a. Sketch the distribution of the variable $x$.
b. Obtain the standardized version, $z$, of $x$.
c. Identify and sketch the distribution of $z$.
d. The area under the normal curve with parameters 18.8 and 1.1 that lies between 17 and 20 is 0.8115. Determine the probability that a randomly selected man will have a forearm length between 17 inches and 20 inches.

e. The percentage of men who have forearm length less than 16 inches equals the area under the standard normal curve that lies to the--------- of--------.

Step-by-Step Solution

Answer

(a) The distribution of the variable $x$ as:

(b) The standardized version is $z=\frac{x-18.8}{1.1}$.

(c) The distribution of $z$ as:

(d) The probability of a randomly picked male with a forearm length of $17$ to $20$ inches is $0.8115$.

(e) The left-hand side of the standard normal curve equals the proportion of forearm length less than $16$ inches.

Part (a) Step 1: Given information

To sketch the distribution of the variable.

Part (a) Step 2: Explanation

The length of a man's forearm with$\mu =18.8$ inches and $\sigma =1.1$ inches, $\left(x\right)$ follows a normal distribution.
The empirical rule can be sketched out as follows:

Part (a) Step 3: Explanation

The graph of $x$ below can be obtained by substituting the parameters with values as follows:

Part (b) Step 1: Given information

To obtain the standardized version,$z$ of $x$.

Part (b) Step 2: Explanation

The standard version is $x$ divided by the standard deviation and decreased by the mean:
$z=\frac{x-\mu }{\sigma }$

$=\frac{x-18.8}{1.1}$

As a result, the standardized version $z=\frac{x-18.8}{1.1}$.

Part (c) Step 1: Given information

To identify and sketch the distribution of $z$.

Part (c) Step 2: Explanation

The length of a man's forearm, because $x$ has a normal distribution, the standardized form of $x$ will also have a normal distribution.
Let, the standard normal value of $15.5$ is,
$z=\frac{15.5-18.8}{1.1}$

$=-3$

Since, $16.6,17.7,18.8,19.9,21$, and $22.1$have standard normal values of $-2,-1,0,1,2$, and $3$ correspondingly.

The distribution of $z$ is depicted in the graph below.

Part (d) Step 1: Given information

To determine the probability that a randomly selected man will have a forearm length between $17$ inches and $20$ inches.

Part (d) Step 2: Explanation

The $0.8115$ portion of area between $17$ and $20$ is found in a normal distribution with parameters of $18.8$ and $1.1$.
The probability inside a certain range is equal to the area under the normal curve in that range.
As a result, the probability of a randomly picked male with a forearm length of $17$ to $20$ inches is $0.8115$.

Part (e) Step 1: Given information

To find the percentage of men who have forearm length less than $16$ inches equals the area under the standard normal curve that lies.

Part (e) Step 2: Explanation

The $z$ score of $16$ is calculated as:
$z=\frac{16-18.8}{1.1}$

$\approx -2.55$

The probability inside a particular range is equal to the area under the standard normal curve in that range.

As a result, the left-hand side of the standard normal curve equals the proportion of forearm length less than $16$ inches.