Question
A lake contains a type of fish called carp. The lengths, $L$ cm, of the carp can be modelled by a normal distribution with mean 45.6 cm and standard deviation 4.2 cm.
According to this model, carp with a length between 41.4 cm and $k$ cm lie within one standard deviation of the mean.
(a) Write down the value of $k$.
(b) Find the probability that a randomly selected carp is greater than 48 cm in length.
(c) It is known that 99% of carp in the lake have a length greater than $x$ cm. Find the value of $x$.
(d) Consider a random sample of 100 carp from the lake.
(i) Find the expected number of carp with lengths between 40 cm and 56 cm.
(ii) Find the probability that in this sample, exactly 95 carp have a length between 40 cm and 56 cm.
A large sample of carp from the lake is studied. The length of each fish is measured and recorded correct to the nearest 0.1 cm.
(e) Find the probability that a randomly selected carp has a length recorded as 45.6 cm.
▶️Answer/Explanation
Detail Solution
part(a)
(a) Write down the value of \(k\).
Step 1: Identify the range of lengths within one standard deviation of the mean.
The mean length of the carp is \(45.6\) cm, and the standard deviation is \(4.2\) cm. One standard deviation below the mean is calculated as:
\(L_{min} = 45.6 – 4.2 = 41.4\) cm.
Step 2: Calculate the upper limit \(k\) using one standard deviation above the mean.
\(L_{max} = 45.6 + 4.2 = 49.8\) cm.
Thus, \(k = 49.8\) cm.
The answer is: 49.8
part (b) Find the probability that a randomly selected carp is greater than 48 cm in length.**
Step 1: Standardize the length using the Z-score formula.
The Z-score is calculated as:
\(Z = \frac{X – \mu}{\sigma} = \frac{48 – 45.6}{4.2} = \frac{2.4}{4.2} \approx 0.5714\).
Step 2: Use the Z-table to find the probability corresponding to \(Z = 0.5714\).
The probability \(P(Z < 0.5714) \approx 0.7157\).
Step 3: Calculate the probability that a carp is greater than 48 cm.
\(P(X > 48) = 1 – P(Z < 0.5714) = 1 – 0.7157 = 0.2843\).
The answer is: 0.2843
(c) Find the value of \(x\) such that 99% of carp have a length greater than \(x\) cm.
Step 1: Identify the Z-score that corresponds to the 1% in the lower tail.
From Z-tables, the Z-score for the lower 1% is approximately \(-2.33\).
Step 2: Use the Z-score to find \(x\).
Using the Z-score formula:
\(-2.33 = \frac{x – 45.6}{4.2}\).
Step 3: Rearrange to solve for \(x\).
\(x – 45.6 = -2.33 \times 4.2\)
\(x = 45.6 – 9.786 = 35.814\).
The answer is: 35.814
(d)(i) Find the expected number of carp with lengths between 40 cm and 56 cm.
Step 1: Calculate the Z-scores for 40 cm and 56 cm.
For \(40\) cm:
\(Z_{40} = \frac{40 – 45.6}{4.2} = \frac{-5.6}{4.2} \approx -1.3333\).
For \(56\) cm:
\(Z_{56} = \frac{56 – 45.6}{4.2} = \frac{10.4}{4.2} \approx 2.4762\).
Step 2: Find the probabilities from the Z-table.
\(P(Z < -1.3333) \approx 0.0918\) and \(P(Z < 2.4762) \approx 0.9934\).
Step 3: Calculate the probability of lengths between 40 cm and 56 cm.
\(P(40 < X < 56) = P(Z < 2.4762) – P(Z < -1.3333) = 0.9934 – 0.0918 = 0.9016\).
Step 4: Multiply by the sample size to find the expected number.
Expected number = \(100 \times 0.9016 = 90.16\).
The answer is: 90.16
(ii) Find the probability that in this sample, exactly 95 carp have a length between 40 cm and 56 cm.
This is a binomial probability: \( X \sim \text{Bin}(100, 0.9016) \), and we need \( P(X = 95) \).
– \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
– \( p = 0.9016 \), \( 1 – p = 0.0984 \), \( n = 100 \), \( k = 95 \).
– \( P(X = 95) = \binom{100}{95} (0.9016)^{95} (0.0984)^{5} \).
– \( \binom{100}{95} = \binom{100}{5} = \frac{100 \cdot 99 \cdot 98 \cdot 97 \cdot 96}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 75,287,520 / 120 = 752,876 \).
Computing \( (0.9016)^{95} \) and \( (0.0984)^5 \) directly is complex, so we use the normal approximation since \( n \) is large:
– Mean: \( \mu = n p = 100 \times 0.9016 = 90.16 \).
– Variance: \( \sigma^2 = n p (1-p) = 100 \times 0.9016 \times 0.0984 \approx 8.875 \).
– Standard deviation: \( \sigma \approx \sqrt{8.875} \approx 2.979 \).
For \( X = 95 \), apply continuity correction (since binomial is discrete):
– \( P(X = 95) \approx P(94.5 < Y < 95.5) \), where \( Y \sim N(90.16, 2.979^2) \).
– Z-scores:
– \( z_1 = \frac{94.5 – 90.16}{2.979} \approx 1.458 \)
– \( z_2 = \frac{95.5 – 90.16}{2.979} \approx 1.794 \).
– \( P(1.458 < Z < 1.794) = P(Z < 1.794) – P(Z < 1.458) \approx 0.9633 – 0.9279 = 0.0354 \).
So, the probability is approximately 0.035 (to 3 decimal places).
(e) Find the probability that a randomly selected carp has a length recorded as 45.6 cm.
Lengths are recorded to the nearest 0.1 cm, so “recorded as 45.6 cm” means the actual length \( L \) is between 45.55 cm and 45.65 cm (since 45.55 rounds up to 45.6, and 45.65 rounds down to 45.6).
– \( P(45.55 < L < 45.65) \).
– Z-scores:
– \( z_1 = \frac{45.55 – 45.6}{4.2} = \frac{-0.05}{4.2} \approx -0.0119 \approx -0.012 \)
– \( z_2 = \frac{45.65 – 45.6}{4.2} = \frac{0.05}{4.2} \approx 0.0119 \approx 0.012 \).
– \( P(-0.012 < Z < 0.012) = P(Z < 0.012) – P(Z < -0.012) \).
Since \( z \) is small, \( P(Z < 0.012) \approx 0.5048 \), and by symmetry, \( P(Z < -0.012) \approx 0.4952 \).
– \( P = 0.5048 – 0.4952 = 0.0096 \).
Alternatively, for small intervals near the mean, use the probability density function approximation:
– PDF at \( \mu = 45.6 \): \( f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-0} = \frac{1}{4.2 \sqrt{2\pi}} \approx 0.0951 \).
– Interval width = 0.1 cm, so \( P \approx 0.0951 \times 0.1 = 0.00951 \).
Both methods agree, so the probability is approximately 0.0095 (to 4 decimal places).
————Markscheme—————–
(a) recognition to add $\mu$ and $\sigma$
49.8 (cm)
(b) $P(L > 48)$
= 0.283854…
= 0.284
(c) $P(L > x) = 0.99$ OR $P(L < x) = 0.01$
$x = 35.8293…$
$x = 35.8 (cm)$
(d) (i) $P(40<L<56)=0.902149…$ (may be seen in part (ii))
attempts to find $100 \times P(40<L<56)$ with their probability
= 90.2149…
= 90.2
(ii) recognizes binomial distribution
$X \sim B(100, 0.902149…)$
$P(X=95)=0.038105…$
= 0.0381
(e) $P(45.55 \leq L < 45.65)$
= 0.009498…
= 0.00950