Magic Math Exercises

1Missing Frequencies

The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies \( f_1 \) and \( f_2 \).

Class Interval	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	5	\( f_1 \)	10	\( f_2 \)	7	8

1

Sum of known frequencies:

\[ 5 + 10 + 7 + 8 = 30 \]

Therefore:

\[ f_1 + f_2 = 50 - 30 = 20 \]

2

Mean calculation setup:

\[ \frac{30f_1 + 70f_2 + 2060}{50} = 62.8 \]

\[ 3f_1 + 7f_2 = 108 \]

3

Solving the system:

\[ \begin{cases} f_1 + f_2 = 20 \\ 3f_1 + 7f_2 = 108 \end{cases} \] \[ \Rightarrow f_1 = 8, f_2 = 12 \]

Missing frequencies: \( \boxed{f_1 = 8} \) and \( \boxed{f_2 = 12} \)

2Standard Deviation

The diameter of circles (in mm) drawn in a design are given below.

Diameters	33-36	37-40	41-44	45-48	49-52
Number of circles	15	17	21	22	25

Calculate the standard deviation.

1

Calculating mean:

\[ \bar{x} = \frac{4350}{100} = 43.5 \text{ mm} \]

2

Variance calculation:

\[ \sigma^2 = \frac{192309}{100} - (43.5)^2 = 30.84 \]

3

Standard deviation:

\[ \sigma = \sqrt{30.84} \approx 5.553 \text{ mm} \]

Standard deviation ≈ \( \boxed{5.553 \text{ mm}} \)

3Variance Calculation

The frequency distribution is given below.

\( x \)	\( k \)	2\( k \)	3\( k \)	4\( k \)	5\( k \)	6\( k \)
\( f \)	2	1	1	1	1	1

Given variance is 160, determine the value of \( k \).

1

Total frequency: \( \sum f = 7 \)

2

Calculations:

\[ \sum fx = 22k \]

\[ \sum fx^2 = 92k^2 \]

3

Variance formula:

\[ \frac{92k^2}{7} - \left(\frac{22k}{7}\right)^2 = 160 \]

4

Solving:

\[ 160k^2 = 7840 \]

\[ k^2 = 49 \Rightarrow k = 7 \]

The value of \( k \) is \( \boxed{7} \)

4Temperature Conversion

The standard deviation of temperature data in °C is 5. If converted to °F, what is the variance?

1

Conversion formula:

\[ °F = \frac{9}{5}°C + 32 \]

2

Standard deviation transformation:

\[ \sigma_F = \frac{9}{5} \times \sigma_C = 9 \]

3

Variance calculation:

\[ \sigma_F^2 = 9^2 = 81 \]

Variance in Fahrenheit = \( \boxed{81} \)

5Mean and Standard Deviation

For a distribution with \( \sum(x-5)=3 \), \( \sum(x-5)^2=43 \), and 18 observations, find the mean and standard deviation.

1

Calculating mean:

\[ \sum x = 3 + 5 \times 18 = 93 \]

\[ \text{Mean} = \frac{93}{18} \approx 5.1667 \]

2

Variance calculation:

\[ \sum (x - \bar{x})^2 = 43 - 18 \times \left(\frac{1}{6}\right)^2 = 42.5 \]

\[ \sigma = \sqrt{\frac{42.5}{18}} \approx 1.537 \]

Mean ≈ \( \boxed{5.1667} \), Standard deviation ≈ \( \boxed{1.537} \)

6Price Stability

Compare price stability between two cities:

City A Prices	20	22	19	23	16
City B Prices	10	20	18	12	15

1

City A calculations:

\[ \text{Mean} = 20, \sigma \approx 2.449, CV \approx 12.25\% \]

2

City B calculations:

\[ \text{Mean} = 15, \sigma \approx 3.688, CV \approx 24.59\% \]

City A has more stable prices (lower CV)

7Range Calculation

Given range = 20 and coefficient of range = 0.2, find largest and smallest values.

1

Equations:

\[ L - S = 20 \]

\[ \frac{L - S}{L + S} = 0.2 \]

2

Solving:

\[ S = 40, L = 60 \]

Smallest value = \( \boxed{40} \), Largest value = \( \boxed{60} \)

8Dice Probability

Find probability of getting product 6 or difference 5 when two dice are rolled.

1

Total outcomes: 36

2

Favorable outcomes:

\[ \text{Product 6: } 4 \text{ outcomes} \]

\[ \text{Difference 5: } 2 \text{ outcomes} \]

Probability = \( \boxed{\dfrac{1}{9}} \)

9Girl Probability

Find probability of at least one girl in a two-child family.

1

Sample space: 4 outcomes

2

Favorable: 3 outcomes

Probability = \( \boxed{\dfrac{3}{4}} \)

10Black Balls

Bag has 5 white and some black balls. If P(Black) = 2 × P(White), find number of black balls.

1

Equation setup:

\[ \frac{b}{5+b} = 2 \times \frac{5}{5+b} \]

Number of black balls = \( \boxed{10} \)

11Exam Probability

Given P(English ∩ Tamil) = 0.5, P(Neither) = 0.1, and P(English) = 0.75, find P(Tamil).

1

Using probability formula:

\[ 0.9 = 0.75 + P(T) - 0.5 \]

P(Tamil) = \( \boxed{0.65} \)