**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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