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EmSAT Math Statistics Sample Questions with Answer explanation
EmSAT Math Statistics Sample Questions
Table of Content
Question 1
Identify Statistical data type for the following variable: a medal won at the Olympics (Gold, Silver, Bronze, None)
A) Nominal
B) Ordinal
C) Interval
D) Ration
Answer: A
Definitions:
- Nominal – applies to data that consists of names, labels, or categories
Example: John, Male, Female
- Ordinal – applies to data that can be arranged in order
Example: Happy, very happy, ok
- Interval – applies to data that can be arranged in order. Addition and differences between data values are meaningful
Example: Temperature in Degree, SAT Score
- Ratio – applies to data that can be arranged in order. Ratios of data values are meaningful
Example: Age, Height, Weight
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Question 2
Here are the score of 20 students on an Algebra test. Find the mean of data set
Score | 0 | 20 | 40 | 60 | 80 | 100 |
Frequency | 3 | 1 | 2 | 4 | 8 | 2 |
Answer: 59
Mean (Average) = Sum of score numbers
Total Number of student
= (0 x 3) + ( 20 x 1) + ( 40 x 2 ) + ( 60 x 4 ) + ( 80 x 8 ) + ( 100 x )
20
= 1180
20
= 59
EmSAT Math Statistics Sample Questions with Answer explanation
Question 3
Ahmed and Hamad plays tennis each week. The probability that Ahmed wins the first match is .
What is the probability that Ahmed wins exactly three of next four matches against Hamad.
A)
B)
C)
D)
Answer: A
This question is based on conditional probability. Conditional probability represented by formula
, Where n = 4, k = 3, p = and (1 – p) =
= 4 x ( )3 x ()1 Where,
= 4 x x
=
Question 4
The average rain fall for the years since 2005 is given below
Year | 2005 | 2006 | 2007 | 2008 |
Amount in cm | 1.345 | 1.408 | 1.537 | 1.580 |
In 2010 there was 2.956 cm of rainfall. How much more rain fell than predicted by the table above. Round your answer to the nearest hundredth.
Answer: 1.28
To find the predicted rainfall, find the differences between the each year of rain fall.
Average rainfall increased from 2005 to 2006 is 6.3%
1.408 – 1.345 = 0.063 or 6.3%
Average rainfall increased from 2006 to 2007 is 12.9%
1.537 – 1.408 = 0.129 or 12.9%
Average rainfall increased from 2007 to 2008 is 4.3%
1.580 – 1.537 = 0.043 0r 4.3%
To find predicted rainfall for 2010, consider the middle value in percent change (i.e 6.3%)
1.580 + 0.063(1.580) = 1.6795
Therefore, predicted average rainfall in 2010 according to table is 1.6795
Find the difference between actual rainfall and predicted rainfall in 2010
2.956 – 1.6795 = 1.276
Rounding it to nearest hundred gives value of 1.28
Therefore 1.28 cm is more rainfall predicted by table
Question 5
The height of boys in a grade 10 class are normally distributed with a mean of 168 cm and standard deviation of 2.5 cm. In which range do 95% of the height approximately falls?
A) 160.5 – 168 cm
B) 160.5 – 170.5 cm
C) 163 – 173 cm
D) 163 – 175.5 cm
Answer: C
Range of height fall under 95% is given as
Range = Mean ± 2 (Standard Deviation)
Mean or average = 168 cm
Standerd deviation = 2.5
Therefore, upper range = 168 cm+ 2(2.5)
= 168 + 5
= 173 cm
Therefore, lower range = 168 cm – 2(2.5)
= 168 – 5
= 163 cm
Range in between height falls 95% is 163 cm – 173 cm
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great explanation