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Problem 7: Normal Approximation to the Binomial Distribution

The proportion of residents who frequently donate blood is given as $p = 0.22$. The sample size is $n = 200$.

To use the normal approximation to the binomial, we first calculate the mean $\mu$ and the standard deviation $\sigma$:

• $\mu = np = 200 \times 0.22 = 44$
• $\sigma = \sqrt{np(1 - p)} = \sqrt{200 \times 0.22 \times 0.78} \approx 5.83$

(a) Probability of Exactly 40 Donors

We need to calculate $P(X = 40)$ using the normal approximation, so we apply the continuity correction:

• We calculate the probability between $39.5$ and $40.5$: $Z_1 = \frac{39.5 - 44}{5.83} \approx -0.77$ $Z_2 = \frac{40.5 - 44}{5.83} \approx -0.60$ Using the standard normal table, we find: $P(Z_1 < Z < Z_2) = P(Z < -0.60) - P(Z < -0.77)$ From the Z-table: $P(Z < -0.60) \approx 0.2743, \quad P(Z < -0.77) \approx 0.2206$ Thus, $P(39.5 < X < 40.5) = 0.2743 - 0.2206 = 0.0537$.

(b) Probability of More Than 54 Donors

We need to calculate $P(X > 54)$. Again, applying continuity correction:

• We calculate $P(X > 54.5)$: $Z = \frac{54.5 - 44}{5.83} \approx 1.80$ From the Z-table: $P(Z > 1.80) = 1 - P(Z < 1.80) \approx 1 - 0.9641 = 0.0359$

(c) Probability of At Most 165 Not Donating

The problem asks for the probability that at most 165 residents do not frequently donate, meaning $X \leq 165$ do not donate blood.

Let $Y = 200 - X$ represent the number of people who frequently donate blood. Therefore, $Y \leq 165$, meaning $X \geq 35$. We now calculate $P(X \leq 35)$:

• Using continuity correction, we calculate $P(X \leq 35.5)$: $Z = \frac{35.5 - 44}{5.83} \approx -1.46$ From the Z-table: $P(Z < -1.46) \approx 0.0721$

Problem 8: Normal Distribution of Lunch Expenses

Given:

• Mean $\mu = 15$
• Standard deviation $\sigma = 1.5$

(a) Percentage of Students Spending Less Than RM12.50

We calculate the Z-score for $X = 12.50$: $Z = \frac{12.50 - 15}{1.5} = \frac{-2.5}{1.5} \approx -1.67$ From the Z-table: $P(Z < -1.67) \approx 0.0475$ So, approximately 4.75% of students spend less than RM12.50 for lunch.

(b) Probability that Mean Expenses for a Sample Exceed RM16.75

For a sample of 30 students, the standard error $SE$ of the mean is: $SE = \frac{\sigma}{\sqrt{n}} = \frac{1.5}{\sqrt{30}} \approx 0.274$ Now, we calculate the Z-score for $X = 16.75$: $Z = \frac{16.75 - 15}{0.274} \approx 6.39$ From the Z-table, a Z-score of 6.39 is extremely high, leading to a probability close to 0. Hence, the probability is virtually zero that the mean lunch expense of a sample of 30 students exceeds RM16.75.

Would you like further details on any of these solutions, or do you have any questions?

Here are 5 related questions to expand on these concepts:

1. What is the difference between binomial distribution and normal distribution?
2. How does the continuity correction affect binomial approximations using the normal distribution?
3. What are the applications of normal distribution in real-world problems?
4. How does increasing the sample size affect the standard error of the mean?
5. How can you determine the appropriate sample size to achieve a desired level of precision in estimates?

Tip: Always remember to apply continuity correction when approximating binomial probabilities using the normal distribution.