3 Markah murid Tingkatan 5 Zeta dalam ujian Matematik Tambahan adalah mengikut taburan normal dengan
kear min, dan sisihan piawai, daripada murid tersebut memperoleh kurang daripada 45 markah dan
daripada murid tersebut memperoleh lebih daripada 65 markah.
The marks of Form 5 Zeta students in Additional Mathematics test is normally distributed with mean, , and
standard deviation, of the students obtained less than 45 marks and of the students obtained
more than 65 marks.
(a) Tentukan nilai dan .
Determine the values of and .
(b) Jika terdapat 40 orang murid di dalam kelas Tingkatan 5 Zeta, hitung peratus murid yang memperoleh
markah melebihi .
If there are 40 students in class Form 5 Zeta, calculate the percentage of the students who obtained
marks more than .

Ask by Warner Frazier. in Malaysia

Mar 29,2025

Question

3 Markah murid Tingkatan 5 Zeta dalam ujian Matematik Tambahan adalah mengikut taburan normal dengan
kear min, dan sisihan piawai, daripada murid tersebut memperoleh kurang daripada 45 markah dan
daripada murid tersebut memperoleh lebih daripada 65 markah.
The marks of Form 5 Zeta students in Additional Mathematics test is normally distributed with mean, , and
standard deviation, of the students obtained less than 45 marks and of the students obtained
more than 65 marks.
(a) Tentukan nilai dan .
Determine the values of and .
(b) Jika terdapat 40 orang murid di dalam kelas Tingkatan 5 Zeta, hitung peratus murid yang memperoleh
markah melebihi .
If there are 40 students in class Form 5 Zeta, calculate the percentage of the students who obtained
marks more than .

Ask by Warner Frazier. in Malaysia

Mar 29,2025

UpStudy AI · Accepted Answer

**Let $$X$$ be the marks which are normally distributed with mean $$\mu$$ and standard deviation $$\sigma$$.** ### (a) Determining $$\mu$$ and $$\sigma$$ We are given that: - $$P(X < 45) = 0.30$$ - $$P(X > 65) = 0.25$$ which implies $$P(X < 65) = 0.75$$ Standardise using the $$z$$-score formula: $$ z = \frac{X-\mu}{\sigma} $$ 1. For $$X = 45$$: $$ P\left(Z < \frac{45-\mu}{\sigma}\right) = 0.30 $$ Let $$ \frac{45-\mu}{\sigma} = z_{0.30} $$ From standard normal tables, $$z_{0.30} \approx -0.5244$$. Thus, $$ \frac{45 - \mu}{\sigma} = -0.5244 \quad \Rightarrow \quad 45 - \mu = -0.5244\,\sigma \quad \Rightarrow \quad \mu = 45 + 0.5244\,\sigma \quad \text{(Equation 1)} $$ 2. For $$X = 65$$: $$ P\left(Z < \frac{65-\mu}{\sigma}\right) = 0.75 $$ Let $$ \frac{65-\mu}{\sigma} = z_{0.75} $$ From standard normal tables, $$z_{0.75} \approx 0.6745$$. Hence, $$ \frac{65-\mu}{\sigma} = 0.6745 \quad \Rightarrow \quad 65 - \mu = 0.6745\,\sigma \quad \Rightarrow \quad \mu = 65 - 0.6745\,\sigma \quad \text{(Equation 2)} $$ Now, set the two expressions for $$\mu$$ equal to each other: $$ 45 + 0.5244\,\sigma = 65 - 0.6745\,\sigma $$ Solve for $$\sigma$$: $$ 0.5244\,\sigma + 0.6745\,\sigma = 65 - 45 \quad \Rightarrow \quad 1.1989\,\sigma = 20 $$ $$ \sigma = \frac{20}{1.1989} \approx 16.67 $$ Substitute $$\sigma$$ back into either equation to find $$\mu$$. Using Equation 1: $$ \mu = 45 + 0.5244 \times 16.67 \approx 45 + 8.74 = 53.74 $$ Thus, we have: $$ \mu \approx 53.74 \quad \text{and} \quad \sigma \approx 16.67 $$ ### (b) Calculating the Percentage of Students Obtaining More than 80 Marks We need to find: $$ P(X > 80) $$ Standardise: $$ z = \frac{80 - \mu}{\sigma} = \frac{80 - 53.74}{16.67} \approx \frac{26.26}{16.67} \approx 1.575 $$ Using the standard normal table, find: $$ P(Z < 1.575) \approx 0.9429 $$ Therefore, $$ P(X > 80) = 1 - P(Z < 1.575) \approx 1 - 0.9429 = 0.0571 \quad \text{or} \quad 5.71\% $$ **Summary:** - $$\mu \approx 53.74$$ - $$\sigma \approx 16.67$$ - Approximately $$5.71\%$$ of the students score more than 80 marks. $$ \mu \approx 53.74, \quad \sigma \approx 16.67, \quad \text{and approximately } 5.71\% \text{ of the students scored more than } 80 \text{ marks.} $$

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(a) Determining and

(b) Calculating the Percentage of Students Obtaining More than 80 Marks

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