A lab conducts 1000 trail scattering probability experiment. Nine target marbles were used on a 45.0 cm board. Marble diameter 1.48cm.

Question

If the standard value for the marble diameter is 1.55 cm, what is the percent error in the experimental marble diameter?

How many hits were recorded during the experiment?

Answer 1

Hi Vinny!

 

If the experimental diameter is 1.48 cm, then the percent error from the standard value of 1.55 cm is given by:

 

% err = 100*[(|observed - standard|)/standard]  (where observed refers to the experimental value)

 

If the board if 45 cm across (perpendicular to the direction in which the scattered objects are being fired), then the nine target marbles block a certain part of that width.  Each one blocks a width equal to, presumably, 1.48 cm.  So the total amount of the width of the board blocked by the marbles is 1.48 cm * 9 = 13.32 cm.

 

13.32 cm/45 cm = 0.296 is the fraction of the board width blocked by the marbles, so 29.6% of the 45-cm-wide opening is blocked.

 

So, as an approximation (without knowing what is being fired), we can basically say that if the center of what is being fired hits anywhere in this 29.6% of the opening that is blocked, there will be a "hit" recorded.  If we fire evenly acorss the opening (not concentrating our firings in any one or few positions), we can expect 29.6% of the fired objects to register "hits." 

 

I hope that helps set you on your way.  If there is any information I have missed, just let me know.