A batch of resistors is tested to see how close they come to the manufacturer's specification of 47 ohms. Data is tabulated in bins of 0.2 ohm as follow:


Resistance
(Ohms)
46.0-46.2
46.2-46.4
46.4-46.6
46.6-46.8
46.8-47.0
47.0-47.2
47.2-47.4
47.4-47.6
47.6-47.8
47.8-48.0


Frequency
3
5
6
9
5
6
5
2
3
1



Make a histogram of this data. From your plot, estimate the median resistance. What can you say about the accuracy and the precision of the manufacturer's specified resistance of 47 ohms?

so far I have got

The batch size is 3+5+6+9+5+6+5+2+3+1=45

45/2 = 22.5

The first 4 bars add up to 23 so the median is in the right hand side of the 4th bar which is 46.6-46.8

I think this is right so far but I don't know how to figer out were abouts it is in the 4th bar

any help would be appreciated

Note that the median is given by:

\(\frac{n+1}{2}\)^{th} = \(\frac{45+1}{2}\)^{th} = 23^{th}

If there were only 5 resistors, the middle resistor is the 3rd, and if there were 6 resistors, the middle would be the 3.5th (there should be equal number of objects to each side of the median).

Well, you can estimate the resistance of the resistor in question (which is the ninth resistor of the range of 46.6 - 46.8 ohm).

The class width is 0.2 ohm.

Assuming that the resistances are equally distributed within this class width, the ninth resistor will have a resistance of:

46.6 + \frac{n}{9}(0.2) = 46.6 + \frac{9}{9}(0.2) = 46.8ohm

Thanks for that much appreciated