a) For the median value to be 8.5, the average of the 6th and 7th value must be 8.5. this can only happen if the 6th is 8 and the 7th is 9. There is no 9 in the list so x must be 9.

b) From the GDC, the mean value is 9.33 (3sf)

c) From the GDC, the Stndard deviation is 1.97 (3sf)

*Be aware - make sure your calculator is counting '1' of each of the items in list 1 and not looking for a frequency in list 2.

a) From the GDC, r = -0.925 (3sf)

b) From the GDC, y = -0.402x + 12.3 (3sf)

c) y = -0.402 X 10 + 12.3 = 8.28

a) Because the total is 310 (given in the question) the missing frequency must be 42 to make that so.

b) There are 310 results, the median will be between the 155th and 156th, both of these are in this second category

c) From the GDC making sure that it is set up to read list 2 as a frequency. List one should be the midpoints of the class intervals. In this case, with discrete data, there are 5 possible answers for each interval, so the midpoint is the middle one. In the first case, 0 to 4, the 5 posisble answers are 0, 1, 2, 3, 4, so the midpoint is 2.

a) \(\frac { Total\quad 71\quad -\quad 80 }{ Total\quad people } \times Total\quad >1\quad language\\ \frac { 31 }{ 137 } \times 60=13.57664234=13.6\quad (3sf)\)

b)\(\frac { Total\quad 91\quad -\quad 100 }{ Total\quad people } \times Total\quad 1\quad language\\ \frac { 35 }{ 137 } \times 77=19.67153285=19.7\quad (3sf)\)

c) \(dof=(rows-1)(columns-1)\\ =(3-1)(2-1)\\ =2\)

d) From GDC, rounded to 3sf

e) We accept because the probability of independence (p number) is 19% which is higher than the significance level of 10% (this explanation will be expected)

a) From GDC entering

\(Lower\quad limit\quad =\quad 160\\ Upper\quad limit\quad =\quad { 10 }^{ 99 }\quad (or\quad similar\quad large\quad number)\\ \mu \quad =\quad 170\\ \sigma \quad =\quad 6\\ p=0.952\)

b) From GDC entering

\(Lower\quad limit\quad =\quad 180\\ Upper\quad limit\quad =\quad 190\quad \\ \mu \quad =\quad 170\\ \sigma \quad =\quad 6\\ p=0.0473\)

c) Using the inverse normal calculator

\(Tail\quad Left\quad and\quad Area\quad (p)\quad =0.7\quad OR\\ Tail\quad right\quad and\quad area\quad (p)\quad =\quad 0.3\\ \mu \quad =\quad 178\\ \sigma \quad =\quad 8\\ Height\quad =\quad 182.195204\\ =182cm\quad (3sf)\)

d) Two steps

step 1 - Work out the probability of being less than 180cm

\(Lower\quad limit\quad =\quad 0\\ Upper\quad limit\quad =\quad 180\quad \\ \mu \quad =\quad 178\\ \sigma \quad =\quad 6\\ p=0.59870632=0.599\quad (3sf)\)

Now you know that the probability of being less than 'y' = 0.599-0.3=0.299, because the 0.3 area is between y and 180, so, using inverse normal

\(Tail\quad Left\quad and\quad Area\quad (p)\quad =0.299\quad \\ \mu \quad =\quad 178\\ \sigma \quad =\quad 8\\ Height\quad =\quad 173.78177\\ =174cm\quad (3sf)\)