Question
Answer the questions below. Show your solutions by writing step by step.
QUESTION THREE
a) Family income is normally distributed with mean k25000 and standard deviation of k4000. If the poverty level income is below k15000, calculate percentage of the population which lives in poverty.
b) The revenue function of a certain product is given as;
Rev(q) = 2750q – 5q2
Where q = number of units produced and sold
(i) Find the quantity that will maximise revenue.
(ii) Find the maximum revenue.
C ) A company offers annual payment of k1000 at the end of each year for the next three years. Find the present value of this annuity discounted at 7% interest per annum.
QUESTION FOUR
a) It is estimated that 55% of students that are entering ZIPS programme are certain to graduate in 3 years time. From a random sample of six students;
(i) What is the probability that exactly three will graduate?
(ii) What is the probability that more than four will graduate?
(iii) What is the mean number of students that would graduate from a group of six?
b) Find the compound interest rate required for k60000 to amount to k150000 in 5 years.
c) The average price of a maths textbook is k450 and standard deviation of k55. The prices are normally distributed. Calculate the probability of a book with price.
(i) Between mean price and k518.75
(ii) Between k461 and k577.05
QUESTION THREE
a) Family income is normally distributed with mean k25000 and standard deviation of k4000. If the poverty level income is below k15000, calculate percentage of the population which lives in poverty.
b) The revenue function of a certain product is given as;
Rev(q) = 2750q – 5q2
Where q = number of units produced and sold
(i) Find the quantity that will maximise revenue.
(ii) Find the maximum revenue.
C ) A company offers annual payment of k1000 at the end of each year for the next three years. Find the present value of this annuity discounted at 7% interest per annum.
QUESTION FOUR
a) It is estimated that 55% of students that are entering ZIPS programme are certain to graduate in 3 years time. From a random sample of six students;
(i) What is the probability that exactly three will graduate?
(ii) What is the probability that more than four will graduate?
(iii) What is the mean number of students that would graduate from a group of six?
b) Find the compound interest rate required for k60000 to amount to k150000 in 5 years.
c) The average price of a maths textbook is k450 and standard deviation of k55. The prices are normally distributed. Calculate the probability of a book with price.
(i) Between mean price and k518.75
(ii) Between k461 and k577.05
Answers
GPT-4o mini
Let's solve each part step by step.
### QUESTION THREE
#### a) Poverty Level Calculation
1. **Given Data**:
- Mean (μ) = K25000
- Standard Deviation (σ) = K4000
- Poverty level = K15000
2. **Calculate the z-score**:
\[
z = \frac{X - \mu}{σ} = \frac{15000 - 25000}{4000} = \frac{-10000}{4000} = -2.5
\]
3. **Look up the z-score**:
Using the z-table, find the percentage of the population below a z-score of -2.5. The table gives the area to the left of z:
\[
P(Z < -2.5) \approx 0.0062 \text{ (or 0.62%)}
\]
4. **Conclusion**:
The percentage of the population that lives in poverty is approximately **0.62%**.
#### b) Revenue Function
**(i) Find the quantity that will maximize revenue**:
1. **Revenue function**:
\[
Rev(q) = 2750q - 5q^2
\]
2. **Differentiate Rev(q)**:
\[
\frac{dRev}{dq} = 2750 - 10q
\]
3. **Set the derivative to 0 to find critical points**:
\[
2750 - 10q = 0 \Rightarrow 10q = 2750 \Rightarrow q = 275
\]
4. **Conclusion**:
The quantity that will maximize revenue is **275 units**.
**(ii) Find the maximum revenue**:
1. **Substitute q = 275 back into the revenue function**:
\[
Rev(275) = 2750(275) - 5(275^2)
\]
\[
= 756250 - 5(75625) = 756250 - 378125 = 378125
\]
2. **Conclusion**:
The maximum revenue is **K378125**.
#### c) Present Value of Annuity
1. **Given Data**:
- Annual payment (PMT) = K1000
- Number of years (n) = 3
- Interest rate (r) = 7% or 0.07
2. **Formula for Present Value of Annuity**:
\[
PV = PMT \times \frac{1 - (1 + r)^{-n}}{r}
\]
3. **Substitute in values**:
\[
PV = 1000 \times \frac{1 - (1 + 0.07)^{-3}}{0.07}
\]
\[
= 1000 \times \frac{1 - (1.225043)^{-1}}{0.07}
\]
\[
= 1000 \times \frac{1 - 0.815...}{0.07}
\]
\[
= 1000 \times \frac{0.184...}{0.07}
\]
\[
= 1000 \times 2.635... \approx 2635.08
\]
4. **Conclusion**:
The present value of this annuity is approximately **K2635.08**.
### QUESTION FOUR
#### a) Probability of Graduating
**(i) What is the probability that exactly three will graduate?**
1. **Given**: p = 0.55 (probability of graduation), n = 6, k = 3.
2. **Use the binomial probability formula**:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
\[
= \binom{6}{3} (0.55)^3 (0.45)^3
\]
\[
= 20 \times (0.55)^3 \times (0.45)^3
\]
\[
= 20 \times 0.166375 \times 0.091125
\]
\[
= 20 \times 0.015147 = 0.30294 \approx 0.3029
\]
3. **Conclusion**:
The probability that exactly three will graduate is approximately **0.3029**.
**(ii) What is the probability that more than four will graduate?**
1. **Calculate P(X > 4)**:
\[
P(X > 4) = P(X = 5) + P(X = 6)
\]
2. **Calculate P(X = 5)**:
\[
P(X = 5) = \binom{6}{5} (0.55)^5 (0.45)^1
\]
\[
= 6 \times (0.55^5) \times (0.45^1)
\]
\[
= 6 \times 0.0501 \times 0.45 ≈ 0.135135
\]
3. **Calculate P(X = 6)**:
\[
P(X = 6) = \binom{6}{6} (0.55)^6 = 1 \times (0.55^6) = 0.027
\]
4. **Total**:
\[
P(X > 4) = P(X = 5) + P(X = 6) ≈ 0.135135 + 0.027 = 0.162135
\]
5. **Conclusion**:
The probability that more than four will graduate is approximately **0.1621**.
**(iii) Mean number of students that would graduate**:
1. **Mean (μ)** for binomial distribution:
\[
μ = n \times p = 6 \times 0.55 = 3.3
\]
2. **Conclusion**:
The mean number of students that would graduate from a group of six is **3.3**.
#### b) Compound Interest Calculation
1. **Given Data**:
- Present Value (PV) = K60000
- Future Value (FV) = K150000
- n = 5 years
2. **Use the formula for compound interest**:
\[
FV = PV(1 + r)^n
\]
\[
150000 = 60000(1 + r)^5
\]
\[
\frac{150000}{60000} = (1 + r)^5
\]
\[
2.5 = (1 + r)^5
\]
3. **Solve for r**:
\[
(1 + r) = 2.5^{1/5}
\]
Using a calculator, \(2.5^{1/5} \approx 1.2214\).
\[
r = 1.2214 - 1 = 0.2214 \text{ or } 22.14\%
\]
4. **Conclusion**:
The compound interest rate required for K60000 to amount to K150000 in 5 years is approximately **22.14%**.
#### c) Probability of Math Textbook Prices
**(i) Between mean price and K518.75**:
1. **Given Data**:
- Mean (μ) = K450
- SD (σ) = K55
- Calculate the z-scores:
\[
z_{450} = \frac{450 - 450}{55} = 0
\]
\[
z_{518.75} = \frac{518.75 - 450}{55} = \frac{68.75}{55} \approx 1.25
\]
2. **Find probabilities**:
- From z-table:
- \(P(Z < 0) = 0.5000\) (50%)
- \(P(Z < 1.25) \approx 0.8944\) (approximately 89.44%)
3. **Calculate probability between**:
\[
P(450 < X < 518.75) = P(Z < 1.25) - P(Z < 0) = 0.8944 - 0.5000 \approx 0.3944
\]
4. **Conclusion**:
The probability of a book with price between the mean price and K518.75 is approximately **39.44%**.
**(ii) Between K461 and K577.05**:
1. **Calculate z-scores**:
\[
z_{461} = \frac{461 - 450}{55} \approx 0.2
\]
\[
z_{577.05} = \frac{577.05 - 450}{55} \approx 2.31
\]
2. **Find probabilities**:
- From z-table:
- \(P(Z < 0.2) \approx 0.5793\)
- \(P(Z < 2.31) \approx 0.9895\)
3. **Calculate probability between**:
\[
P(461 < X < 577.05) = P(Z < 2.31) - P(Z < 0.2) = 0.9895 - 0.5793 \approx 0.4102
\]
4. **Conclusion**:
The probability of a book with price between K461 and K577.05 is approximately **41.02%**.
### Final Remarks
This had a lot of parts, but we have calculated everything step-by-step. If you need further clarification on any of the steps, feel free to ask!
### QUESTION THREE
#### a) Poverty Level Calculation
1. **Given Data**:
- Mean (μ) = K25000
- Standard Deviation (σ) = K4000
- Poverty level = K15000
2. **Calculate the z-score**:
\[
z = \frac{X - \mu}{σ} = \frac{15000 - 25000}{4000} = \frac{-10000}{4000} = -2.5
\]
3. **Look up the z-score**:
Using the z-table, find the percentage of the population below a z-score of -2.5. The table gives the area to the left of z:
\[
P(Z < -2.5) \approx 0.0062 \text{ (or 0.62%)}
\]
4. **Conclusion**:
The percentage of the population that lives in poverty is approximately **0.62%**.
#### b) Revenue Function
**(i) Find the quantity that will maximize revenue**:
1. **Revenue function**:
\[
Rev(q) = 2750q - 5q^2
\]
2. **Differentiate Rev(q)**:
\[
\frac{dRev}{dq} = 2750 - 10q
\]
3. **Set the derivative to 0 to find critical points**:
\[
2750 - 10q = 0 \Rightarrow 10q = 2750 \Rightarrow q = 275
\]
4. **Conclusion**:
The quantity that will maximize revenue is **275 units**.
**(ii) Find the maximum revenue**:
1. **Substitute q = 275 back into the revenue function**:
\[
Rev(275) = 2750(275) - 5(275^2)
\]
\[
= 756250 - 5(75625) = 756250 - 378125 = 378125
\]
2. **Conclusion**:
The maximum revenue is **K378125**.
#### c) Present Value of Annuity
1. **Given Data**:
- Annual payment (PMT) = K1000
- Number of years (n) = 3
- Interest rate (r) = 7% or 0.07
2. **Formula for Present Value of Annuity**:
\[
PV = PMT \times \frac{1 - (1 + r)^{-n}}{r}
\]
3. **Substitute in values**:
\[
PV = 1000 \times \frac{1 - (1 + 0.07)^{-3}}{0.07}
\]
\[
= 1000 \times \frac{1 - (1.225043)^{-1}}{0.07}
\]
\[
= 1000 \times \frac{1 - 0.815...}{0.07}
\]
\[
= 1000 \times \frac{0.184...}{0.07}
\]
\[
= 1000 \times 2.635... \approx 2635.08
\]
4. **Conclusion**:
The present value of this annuity is approximately **K2635.08**.
### QUESTION FOUR
#### a) Probability of Graduating
**(i) What is the probability that exactly three will graduate?**
1. **Given**: p = 0.55 (probability of graduation), n = 6, k = 3.
2. **Use the binomial probability formula**:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
\[
= \binom{6}{3} (0.55)^3 (0.45)^3
\]
\[
= 20 \times (0.55)^3 \times (0.45)^3
\]
\[
= 20 \times 0.166375 \times 0.091125
\]
\[
= 20 \times 0.015147 = 0.30294 \approx 0.3029
\]
3. **Conclusion**:
The probability that exactly three will graduate is approximately **0.3029**.
**(ii) What is the probability that more than four will graduate?**
1. **Calculate P(X > 4)**:
\[
P(X > 4) = P(X = 5) + P(X = 6)
\]
2. **Calculate P(X = 5)**:
\[
P(X = 5) = \binom{6}{5} (0.55)^5 (0.45)^1
\]
\[
= 6 \times (0.55^5) \times (0.45^1)
\]
\[
= 6 \times 0.0501 \times 0.45 ≈ 0.135135
\]
3. **Calculate P(X = 6)**:
\[
P(X = 6) = \binom{6}{6} (0.55)^6 = 1 \times (0.55^6) = 0.027
\]
4. **Total**:
\[
P(X > 4) = P(X = 5) + P(X = 6) ≈ 0.135135 + 0.027 = 0.162135
\]
5. **Conclusion**:
The probability that more than four will graduate is approximately **0.1621**.
**(iii) Mean number of students that would graduate**:
1. **Mean (μ)** for binomial distribution:
\[
μ = n \times p = 6 \times 0.55 = 3.3
\]
2. **Conclusion**:
The mean number of students that would graduate from a group of six is **3.3**.
#### b) Compound Interest Calculation
1. **Given Data**:
- Present Value (PV) = K60000
- Future Value (FV) = K150000
- n = 5 years
2. **Use the formula for compound interest**:
\[
FV = PV(1 + r)^n
\]
\[
150000 = 60000(1 + r)^5
\]
\[
\frac{150000}{60000} = (1 + r)^5
\]
\[
2.5 = (1 + r)^5
\]
3. **Solve for r**:
\[
(1 + r) = 2.5^{1/5}
\]
Using a calculator, \(2.5^{1/5} \approx 1.2214\).
\[
r = 1.2214 - 1 = 0.2214 \text{ or } 22.14\%
\]
4. **Conclusion**:
The compound interest rate required for K60000 to amount to K150000 in 5 years is approximately **22.14%**.
#### c) Probability of Math Textbook Prices
**(i) Between mean price and K518.75**:
1. **Given Data**:
- Mean (μ) = K450
- SD (σ) = K55
- Calculate the z-scores:
\[
z_{450} = \frac{450 - 450}{55} = 0
\]
\[
z_{518.75} = \frac{518.75 - 450}{55} = \frac{68.75}{55} \approx 1.25
\]
2. **Find probabilities**:
- From z-table:
- \(P(Z < 0) = 0.5000\) (50%)
- \(P(Z < 1.25) \approx 0.8944\) (approximately 89.44%)
3. **Calculate probability between**:
\[
P(450 < X < 518.75) = P(Z < 1.25) - P(Z < 0) = 0.8944 - 0.5000 \approx 0.3944
\]
4. **Conclusion**:
The probability of a book with price between the mean price and K518.75 is approximately **39.44%**.
**(ii) Between K461 and K577.05**:
1. **Calculate z-scores**:
\[
z_{461} = \frac{461 - 450}{55} \approx 0.2
\]
\[
z_{577.05} = \frac{577.05 - 450}{55} \approx 2.31
\]
2. **Find probabilities**:
- From z-table:
- \(P(Z < 0.2) \approx 0.5793\)
- \(P(Z < 2.31) \approx 0.9895\)
3. **Calculate probability between**:
\[
P(461 < X < 577.05) = P(Z < 2.31) - P(Z < 0.2) = 0.9895 - 0.5793 \approx 0.4102
\]
4. **Conclusion**:
The probability of a book with price between K461 and K577.05 is approximately **41.02%**.
### Final Remarks
This had a lot of parts, but we have calculated everything step-by-step. If you need further clarification on any of the steps, feel free to ask!