Question

Answer the questions below. Show your solutions by writing step by step.

QUESTION ONE
a) A company has the following production figures for the month of January 2019.

Weight range frequency
10 ˂ 20 10
20 ˂ 30 11
30 ˂ 40 14
40 ˂ 50 13

Calculate
(i) The mean weight.

(ii) The standard deviation.


b) Mr. Thambo deposited k1500 in an account that pays 12% compound interest per annum. How long will it take for it to amount to k2500?

c) Solve the equation using factorization. 3X2 + 4x – 15 = 0 (5 marks)
Total 20 marks

Answer 1

Let's go through each part of the question step-by-step.

### QUESTION ONE

#### a) A company has the following production figures for the month of January 2019.
Weight range:
- 10 ˂ 20: Frequency = 10
- 20 ˂ 30: Frequency = 11
- 30 ˂ 40: Frequency = 14
- 40 ˂ 50: Frequency = 13

We need to calculate:
- (i) The mean weight
- (ii) The standard deviation

##### (i) Calculate the mean weight

1. **Find the midpoint of each weight range:**
- \( 10 ˂ 20 \): Midpoint = \( \frac{10 + 20}{2} = 15 \)
- \( 20 ˂ 30 \): Midpoint = \( \frac{20 + 30}{2} = 25 \)
- \( 30 ˂ 40 \): Midpoint = \( \frac{30 + 40}{2} = 35 \)
- \( 40 ˂ 50 \): Midpoint = \( \frac{40 + 50}{2} = 45 \)

2. **Multiply each midpoint by its respective frequency:**
- \( 15 \times 10 = 150 \)
- \( 25 \times 11 = 275 \)
- \( 35 \times 14 = 490 \)
- \( 45 \times 13 = 585 \)

3. **Sum the products:**
\[
150 + 275 + 490 + 585 = 1500
\]

4. **Sum the frequencies:**
\[
10 + 11 + 14 + 13 = 48
\]

5. **Calculate the mean:**
\[
\text{Mean} = \frac{\text{Total of midpoints}}{\text{Total frequency}} = \frac{1500}{48} \approx 31.25
\]

##### (ii) Calculate the standard deviation

1. **Calculate the variance using the formula:**
\[
\sigma^2 = \frac{\sum f(x - \text{mean})^2}{N}
\]
where \( f \) is the frequency, \( x \) is the midpoint, and \( N \) is the total frequency.

2. **Calculate \( (x - \text{mean})^2 \) for each midpoint:**
- For \( 15 \): \( (15 - 31.25)^2 = (16.25)^2 = 264.0625 \)
- For \( 25 \): \( (25 - 31.25)^2 = (6.25)^2 = 39.0625 \)
- For \( 35 \): \( (35 - 31.25)^2 = (3.75)^2 = 14.0625 \)
- For \( 45 \): \( (45 - 31.25)^2 = (13.75)^2 = 187.5625 \)

3. **Multiply each result by the frequency:**
- \( 264.0625 \times 10 = 2640.625 \)
- \( 39.0625 \times 11 = 429.6875 \)
- \( 14.0625 \times 14 = 196.875 \)
- \( 187.5625 \times 13 = 2438.3125 \)

4. **Sum these values:**
\[
2640.625 + 429.6875 + 196.875 + 2438.3125 = 5730.5
\]

5. **Calculate the variance:**
\[
\sigma^2 = \frac{5730.5}{48} \approx 119.765625
\]

6. **Calculate the standard deviation:**
\[
\sigma = \sqrt{119.765625} \approx 10.95
\]

#### b) Mr. Thambo deposited k1500 in an account that pays 12% compound interest per annum. How long will it take for it to amount to k2500?

We use the compound interest formula:
\[
A = P(1 + r)^t
\]
where:
- \( A = 2500 \)
- \( P = 1500 \)
- \( r = 0.12 \)

1. **Set up the equation:**
\[
2500 = 1500(1 + 0.12)^t
\]

2. **Simplify the equation:**
\[
\frac{2500}{1500} = (1.12)^t
\]
\[
\frac{5}{3} \approx 1.6667 = (1.12)^t
\]

3. **Take the natural logarithm (ln) of both sides:**
\[
\ln(1.6667) = t \ln(1.12)
\]
\[
t = \frac{\ln(1.6667)}{\ln(1.12)}
\]

4. **Calculate \( \ln(1.6667) \) and \( \ln(1.12) \):**
\[
\ln(1.6667) \approx 0.5108, \quad \ln(1.12) \approx 0.1133
\]
\[
t \approx \frac{0.5108}{0.1133} \approx 4.51
\]

So, it will take approximately **4.51 years** for the amount to grow from k1500 to k2500.

#### c) Solve the equation using factorization: \( 3x^2 + 4x - 15 = 0 \)

1. **Multiply \( a \) and \( c \) (3 and -15):**
\[
ac = 3 \times -15 = -45
\]

2. **Find two numbers that multiply to \( -45 \) and add up to \( 4 \):**
- The numbers are \( 9 \) and \( -5 \).

3. **Rewrite the equation:**
\[
3x^2 + 9x - 5x - 15 = 0
\]

4. **Group the terms:**
\[
(3x^2 + 9x) + (-5x - 15) = 0
\]
\[
3x(x + 3) - 5(x + 3) = 0
\]

5. **Factor by grouping:**
\[
(3x - 5)(x + 3) = 0
\]

6. **Set each factor to zero:**
- \( 3x - 5 = 0 \) gives \( x = \frac{5}{3} \)
- \( x + 3 = 0 \) gives \( x = -3 \)

Thus, the solutions are \( x = \frac{5}{3} \) and \( x = -3 \).

### Summary of Answers:
- (a) (i) Mean weight: **31.25**
- (a) (ii) Standard deviation: **10.95**
- (b) Time for k1500 to become k2500: **4.51 years**
- (c) Solutions of the equation: **x = -3 and x = \frac{5}{3}**