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# omi67 plz help

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

https://vle.mathswatch.co.uk/images/questions/question1084.png

https://vle.mathswatch.co.uk/images/questions/question1086.png

May 2, 2020

#1
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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   May 2, 2020
#2
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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   May 2, 2020
#3
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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   May 2, 2020