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Super weapons development. The U.S. Army is working with a major defense contractor (not named here for both confidentiality and security reasons) to develop a 鈥渟uper鈥� weapon. The weapon is designed to fire a large number of sharp tungsten bullets鈥攃alled flechettes鈥攚ith a single shot that will destroy a large number of enemy soldiers. (Flechettes are about the size of an average nail, with small fins at one end to stabilize them in flight.) The defense contractor has developed a prototype gun that fires 1,100 flechettes with a single round. In range tests, three 2-feet-wide targets were set up at a distance of 500 meters (approximately 1500 feet) from the weapon. Using a number line as a reference, the centers of the three targets were at 0, 5, and 10 feet, respectively, as shown in the accompanying figure. The prototype gun was aimed at the middle target (center at 5 feet) and fired once. The point x where each of the 1,100 flechettes landed at the 500-meter distance was measured using a horizontal grid. For example, a flechette with a horizontal value of x = 5.5 (shown in the figure) hit the middle target, but a flechette with a horizontal value of x = 2.0 (also shown in the figure) did not hit any of the three targets. The 1,100 measurements on the random variable x are saved in the accompanying file. (The data are simulated for confidentiality reasons.) The defense contractor is interested in the likelihood of any one of the targets being hit by a flechette and in particular, wants to set the gun specifications to maximize the number of target hits. The weapon is designed to have a mean horizontal value, $\mathbf{E}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, equal to the aim point (e.g., m = 5 feet when aimed at the center target). By changing specifications, the contractor can vary the standard deviation, role="math" localid="1660730004191" $\mathit{蟽}$. The data file contains flechette measurements for three different range tests鈥攐ne with a standard deviation of $\mathit{蟽}$ = 1 foot, one with $\mathit{蟽}$ = 2 feet, and one with $\mathit{蟽}$= 4 feet. Let ,${\mathit{X}}_{\mathbf{1}}$ ,${\mathit{x}}_{\mathbf{2}}$ and${\mathit{X}}_{\mathbf{4}}$ represent the random variables for horizontal measurements with $\mathit{蟽}$ = 1, $\mathit{蟽}$= 2, and $\mathit{蟽}$= 4, respectively. From past experience, the defense contractor has found that the distribution of the horizontal flechette measurements is closely approximated by a normal distribution.

a. For each of the three values of s, use the normal distribution to find the approximate probability that a single flechette shot from the weapon will hit any one of the three targets. [Hint: Note that the three targets range from -1 to 1, 4 to 6, and 9 to 11 feet on the horizontal grid.]

b. The actual results of the three range tests are saved in the data file. Use this information to calculate the proportion of the 1,100 flechettes that actually hit each target鈥攃alled the hit ratio鈥攆or each value of s. How do these actual hit ratios compare with the estimated probabilities of a hit using the normal distribution?

c. If the U.S. Army wants to maximize the chance of hitting the target that the prototype gun is aimed at, what setting should be used for s? If the Army wants to hit multiple targets with a single shot of the weapon, what setting should be used for s?

Step by step solution

01

Given information

The random variable $x_1$ as the horizontal flechette measurement follows the approximately normal distribution with a mean $\left(\mathrm{渭}\right)$as 5 feet and a standard deviation $\left(\mathrm{蟽}\right)$as 1 foot.

The random variable $x_2$ as the horizontal flechette measurement follows an approximately normal distribution with a mean $\left(\mathrm{渭}\right)$as 5 feet and a standard deviation $\left(蟽\right)$as 2 feet.

The random variable $x_3$ as the horizontal flechette measurement follows an approximately normal distribution with a mean $\left(渭\right)$as 5 feet and a standard deviation $\left(蟽\right)$as 4 feet.

02

(a) Calculating the probability

Let the random variable $x_1$as the horizontal flechette measurement follow an approximately normal distribution with a mean $\left(渭\right)$as 5 feet and a standard deviation $\left(蟽\right)$as 1 foot.

Therefore,

$\begin{array}{c}\mathrm{P}(鈭�1鈮�{\mathrm{x}}_1鈮�1)+\mathrm{P}(4鈮�{\mathrm{x}}_1鈮�6)+\mathrm{P}(9鈮�{\mathrm{x}}_1鈮�11)\\ =\mathrm{P}\left(\frac{鈭�1鈭�5}{1}鈮�\mathrm{z}鈮�\frac{1鈭�5}{1}\right)+\mathrm{P}\left(\frac{4鈭�5}{1}鈮�\mathrm{z}鈮�\frac{6鈭�5}{1}\right)+\mathrm{P}\left(\frac{9鈭�5}{1}鈮�\mathrm{z}鈮�\frac{11鈭�5}{1}\right)\\ =\mathrm{P}(鈭�6鈮�\mathrm{z}鈮�鈭�4)+\mathrm{P}(鈭�1鈮�\mathrm{z}鈮�1)+\mathrm{P}(4鈮�\mathrm{z}鈮�6)\\ =0+0.3413+0.3413\end{array}$

The probability that a single flechette shot will hit any one of the three targets when $\mathrm{蟽}=1$is 0.6826.

Now,

Define the random variable $x_2$as the horizontal flechette measurement follows an approximately normal distribution with a mean $\left(渭\right)$as 5 feet and a standard deviation $\left(蟽\right)$as 2 feet.

$\begin{array}{c}\mathrm{P}(鈭�1鈮�{\mathrm{x}}_1鈮�1)+\mathrm{P}(4鈮�{\mathrm{x}}_1鈮�6)+\mathrm{P}(9鈮�{\mathrm{x}}_1鈮�11)\\ =\mathrm{P}\left(\frac{鈭�1鈭�5}{2}鈮�\mathrm{z}鈮�\frac{1鈭�5}{2}\right)+\mathrm{P}\left(\frac{4鈭�5}{2}鈮�\mathrm{z}鈮�\frac{6鈭�5}{2}\right)+\mathrm{P}\left(\frac{9鈭�5}{2}鈮�\mathrm{z}鈮�\frac{11鈭�5}{2}\right)\\ =\mathrm{P}(鈭�3鈮�\mathrm{z}鈮�鈭�2)+\mathrm{P}(鈭�0.5鈮�\mathrm{z}鈮�0.5)+\mathrm{P}(2鈮�\mathrm{z}鈮�3)\\ =(0.4987鈭�0.4772)+(0.1915+0.1915)+(0.4987鈭�0.4772)\end{array}$

$=0.4260$

The probability that a single flechette shot will hit any one of the three targets when $蟽=2$is 0.4260.

Now,

Define the random variable $x_4$ as the horizontal flechette measurement follows the approximately normal distribution with a mean $\left(渭\right)$as 5 feet and a standard deviation $\left(蟽\right)$as 4 foot.

$\begin{array}{c}\mathrm{P}(鈭�1鈮�{\mathrm{x}}_1鈮�1)+\mathrm{P}(4鈮�{\mathrm{x}}_1鈮�6)+\mathrm{P}(9鈮�{\mathrm{x}}_1鈮�11)\\ =\mathrm{P}\left(\frac{鈭�1鈭�5}{4}鈮�\mathrm{z}鈮�\frac{1鈭�5}{4}\right)+\mathrm{P}\left(\frac{4鈭�5}{4}鈮�\mathrm{z}鈮�\frac{6鈭�5}{4}\right)+\mathrm{P}\left(\frac{9鈭�5}{4}鈮�\mathrm{z}鈮�\frac{11鈭�5}{4}\right)\\ =\mathrm{P}(鈭�1.5鈮�\mathrm{z}鈮�鈭�1)+\mathrm{P}(鈭�0.25鈮�\mathrm{z}鈮�0.25)+\mathrm{P}(1鈮�\mathrm{z}鈮�1.5)\\ =(0.4332鈭�0.3413)+(0.0987+0.0987)+(0.4332鈭�0.3413)\end{array}$

$=0.3812$

The probability that a single flechette shot will hit any of the three targets $蟽=4$ is 0.3812.

03

(b) Calculating the proportion

The proportion of flechettes that actually hit the target when $\mathit{蟽}\mathbf{=}\mathbf{1}$:

The number of flechettes that hit the target is 764. So, the proportion of flechettes that hit the target is $\frac{764}{1100}$.

But, the probability that a single flechette shot will hit any of the three targets is 0.6826, which is less than 0.6945.

The proportion of flechettes that actually hit the target whenrole="math" localid="1660731944590" $\mathbf{蟽}\mathbf{=}\mathbf{2}$:

The number of flechettes that hit the target is 462. So, the proportion of flechettes that actually hit the target$\frac{462}{1100}$

But, the probability that a single flechette shot will hit any of the three targets w

The proportion of flechettes that actually hit the target when $\mathbf{蟽}\mathbf{=}\mathbf{4}$:

The number of flechettes that hit the target is 408. So, the proportion of flechettes that actually hit the target $\frac{408}{1100}$

But, the probability that a single flechette shot will hit any of the three targets role="math" $蟽=4$ is 0.3812, which is greater than 0.3709.

04

(c) Result

If the army needs to maximize the chance of hitting the target, then $蟽=1$ is set as the probability is 0.6826.

When the army needs to hit multiple targets with a single shot, then$蟽=2$ is set as the probability is 0.4260.

If the army needs to maximize the chance of hitting the target, then $蟽=4$ is set as the probability is 0.3812.

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