# MDM4U Practice Test (Chapter 3): Solutions ```MDM4U Practice Test (Chapter 3): Answers
1. a) uniform b) skewed left
2. a) mean  7.35 , median=7, mode=7
b)
4(5  7.35) 2  3(6  7.35) 2  8(7  7.35) 2  7(8  7.35) 2  4(9  7.35) 2  2(10  7.35) 2

4  38 7 4  2
56.43
28
 1.42
3. a) discrete b) continuous
4. Favourite Colour
5. Since there are 11 values in each half, Q1 will be the 6th value, and Q3 will be the 17th
value. IQR=Q3-Q1
6. Q1 is 10, so 75% of the values are over 10
7. a) 81.5% of values
b) The mean has a z score of 0
8.
88th percentile correspond s to a z  score of 1.175 ( from table)

So the value is 1.175 standard deviations from the mean.
value  1092  1.175(6)
 1099.05
9. Yes, the distribution is approximately normal.
i)
The mean, median, and mode are very close
ii)
It is bell shaped and symmetrical
iii)
There are approximately 145+190+200+140=675/1000 values within
1 standard deviation of the mean (67.5% - very close to 68%)
10. a) 8,8,8,8,8,8 (all numbers the same)
11.
Calculate z - score for 4000 :
4000 - 5000
400
 1000

400
 2.5
z
From the z - table, P( z  2.5)  0.0062
So 0.62% of days have fewer than 400 hits.
In a year, this correspond s to 0.0062(365)  2.263 days,
So I would expect fewer than 400 hits only 2 days per year.
12. Range=12
Bin width=2
# of bins = 6
bin
frequency
-10
2
12
3
34
4
56
7
78
8
910
2
1112
1
13. Let x represent his exam mark (%)
70(80)  10(72)  20( x )
100
8200  5600  720  20 x
82 
1880  20 x
x  94
Therefore he must get 94% on the exam to get 82% in the course.
14.
z - score for 8th percentile : z  -1.4
 mean - 1.4  12 (1)
z - score for 69th percentile : z  0.5
 mean  0.5  100 (2)
Solving the system of equations :
mean  12  1.4 (1)
mean  100 - 0.5 (2)
Substitute (1) into (2)
12  1.4  100 - 0.5
1.9  88
  46.3
Now solve for mean...
mean  0.5  100 (2)
mean  0.5(46.3)  100
mean  76.85
 The mean rainfall in April is approximat ely 76.85mm
and the standard devaition is approximat ely 46.3 mm
14. Ms. Democracy could have a higher average, but a larger standard deviation (a
very inconsistent class), meaning that she had more failing students.
Mr. Charter could have a very consistent class with a lower average.
For Example: Ms. Democacracy: N (75,252 )  P(X&lt;50) = 15.7% failing
Mr. Charter: N (70,10 2 )  P(X&lt;50) = 2.3% failing
```