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https://vle.mathswatch.co.uk/images/questions/question1075.png

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 May 2, 2020
 #1
avatar+111456 
+1

First one

 

Mean height of team  =

 

[ (1)(198)  + (3)(199)  + (2)(200) + (5)(201)  + (2)(202) ]  / [ 1 + 3 + 2 + 5 + 2 ]  =

 

[2604/ 13]  =

 

200.3 cm

 

 

Let x  be the  height of the new player.....and we  have

 

[ 2604 + x ] / 14   = 201    multiply both sides by 14

 

 

2604 + x   =  2814        subtract 2604 from both sides

 

x  = 210  cm

 

 

cool cool cool

 May 2, 2020
 #2
avatar+111456 
+1

Second one 

 

a) class interval that contains the median

 

We have  30  values

 

The  median  occurs  between the 15th and 16th value

 

The interval where this occurs  is     2 ≤ h < 3

 

b)  mean .....I'm less sure about this one because we don't actually know how many hours any particular individual spends on his/her tablet

 

But let's assume that each individual spends the midpoint of their class interval as a representative time on their tablet

[ For instance....assume that  every one in the first class interval spends   1/2  hr on their phone.....everyone in the second class interval spens 1 + 1/2  hrs on their phone, etc.]

 

 

So......one estimate  of the mean is

 

[ 4 (1/2)  +  8 (1 + 1/2)  + 11(2 + 1/2)  + 7(3 + 1/2) ]   / 30   ≈   2.2 hrs

 

 

cool cool cool

 May 2, 2020
 #3
avatar+111456 
+1

Last one

 

a)  Making the same assumption as in the estimated mean in the  last problem  we have

 

[  2(7.5)  + 9(12.5)  + 5(17.5)  + 5 (22.5)  + 3(27.5) ]  /  [ 2 + 9 + 5 + 5 + 3]  ≈  17  min

 

 

b) Fraction finishing  in less than 20 min   =   [ 5 + 9 + 2 ]   / [ 2 + 9 + 5 + 5 + 3 ]  =  16 / 24  = 2/3  

 

 

cool cool cool

 May 2, 2020

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