Dalam satu kolam menternak ikan air tawar, berat sejenis ikan bertabur secara normal dengan min 1.0 kg dan
varians .
In a fresh water fish pond, the weight of one type of fish is normally distributed with a mean of 1.0 kg and
variance of .
(a) Carikan kebarangkalian seekor ikan yang dipilih secara rawak adalah kurang daripada 0.82 kg .
Find the probability that a fish chosen randomly is less than 0.82 kg .
(b) Diberi daripada ikan mempunyai berat lebih daripada , carikan nilai .
Given that of the fish has the weight more than m kg , find the value of .
Jawapan / Answer.

Ask by Dickson Munoz. in Malaysia

Mar 29,2025

Question

Dalam satu kolam menternak ikan air tawar, berat sejenis ikan bertabur secara normal dengan min 1.0 kg dan
varians .
In a fresh water fish pond, the weight of one type of fish is normally distributed with a mean of 1.0 kg and
variance of .
(a) Carikan kebarangkalian seekor ikan yang dipilih secara rawak adalah kurang daripada 0.82 kg .
Find the probability that a fish chosen randomly is less than 0.82 kg .
(b) Diberi daripada ikan mempunyai berat lebih daripada , carikan nilai .
Given that of the fish has the weight more than m kg , find the value of .
Jawapan / Answer.

Ask by Dickson Munoz. in Malaysia

Mar 29,2025

UpStudy AI · Accepted Answer

$$
\textbf{Diberi:} \quad X \sim N(1.0,\,0.09) \quad \text{jadi} \quad \sigma=\sqrt{0.09}=0.3.
$$

$$
\textbf{(a)} \quad \text{Kebarangkalian} \quad P(X<0.82)
$$
$$
\text{Kira nilai } Z: \quad Z = \frac{0.82 - 1.0}{0.3} = \frac{-0.18}{0.3} = -0.6.
$$
$$
\text{Maka, } P(X<0.82) = P(Z < -0.6).
$$
$$
\text{Menggunakan jadual taburan normal, } P(Z < -0.6) \approx 0.2743.
$$

$$
\textbf{(b)} \quad \text{Diberi bahawa } 30\% \text{ ikan mempunyai berat lebih daripada } m\text{ kg.}
$$
$$
\text{Ini bermakna } P(X>m)=0.30 \quad \text{atau} \quad P(X \le m)=0.70.
$$
$$
\text{Cari } Z \text{ yang memenuhi } P(Z \le z)=0.70.
$$
$$
\text{Daripada jadual taburan, } z \approx 0.524.
$$
$$
\text{Oleh itu, } m = \mu + \sigma \, z = 1.0 + 0.3 \times 0.524 \approx 1.0 + 0.1572 = 1.1572.
$$

$$
\boxed{
\begin{aligned}
\text{(a)}&\ P(X<0.82) \approx 0.2743, \
\text{(b)}&\ m \approx 1.16\text{ kg}.
\end{aligned}
}
$$
(a) The probability that a randomly selected fish weighs less than 0.82 kg is approximately 0.2743. (b) The value of $$ m $$ is approximately 1.16 kg.

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