Calculation of probability is fundamental when making decisions based on data in community health. Health professionals in Kenya often rely on probabilities to assess risks, predict disease outbreaks, and evaluate treatment outcomes. Understanding how to calculate probability helps in interpreting health surveys, clinical trial results, and epidemiological data. This chapter focuses on practical applications of probability in health care delivery using worked examples relevant to community health settings.
Probability measures the likelihood of an event occurring, expressed as a value between 0 and 1. An event with probability 0 is impossible, while 1 indicates certainty. The formula for probability of an event \( A \) is:
$$ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} $$
Example 1: A health worker randomly selects a patient from a clinic with 100 patients, 30 of whom have malaria. What is the probability that the selected patient has malaria?
Given:
Number of favorable outcomes = 30
Total outcomes = 100
$$ P(\text{malaria}) = \frac{30}{100} $$
$$ = 0.3 $$
Answer: 0.3 (30%)
Example 2: In a sample of 120 children, 48 are vaccinated against measles. What is the probability a randomly selected child is not vaccinated?
Given:
Number vaccinated = 48
Total children = 120
Number not vaccinated = 120 - 48 = 72
$$ P(\text{not vaccinated}) = \frac{72}{120} $$
$$ = 0.6 $$
Answer: 0.6 (60%)
Example 3: A community health survey shows that out of 500 households, 125 use clean water sources. What is the probability that a household uses unclean water?
Given:
Households using clean water = 125
Total households = 500
Households using unclean water = 500 - 125 = 375
$$ P(\text{unclean water}) = \frac{375}{500} $$
$$ = 0.75 $$
Answer: 0.75 (75%)
Example 4: In a hospital, the probability that a patient has hypertension is 0.4. What is the probability that a patient selected at random does not have hypertension?
Given:
\( P(\text{hypertension}) = 0.4 \)
$$ P(\text{no hypertension}) = 1 - P(\text{hypertension}) $$
$$ = 1 - 0.4 $$
$$ = 0.6 $$
Answer: 0.6 (60%)
Example 5: A health facility has 200 patients, 50 have diabetes, 30 have tuberculosis, and the rest have other illnesses. What is the probability that a randomly selected patient has diabetes or tuberculosis?
Given:
Patients with diabetes = 50
Patients with tuberculosis = 30
Total patients = 200
$$ P(\text{diabetes or tuberculosis}) = \frac{50 + 30}{200} $$
$$ = \frac{80}{200} $$
$$ = 0.4 $$
Answer: 0.4 (40%)
Two events are independent if the occurrence of one does not affect the probability of the other. The probability of both independent events \( A \) and \( B \) occurring is:
$$ P(A \text{ and } B) = P(A) \times P(B) $$
Example 1: The probability that a patient has malaria is 0.2, and the probability that the same patient has typhoid is 0.1. Assuming these are independent, what is the probability that the patient has both malaria and typhoid?
Given:
\( P(\text{malaria}) = 0.2 \)
\( P(\text{typhoid}) = 0.1 \)
$$ P(\text{malaria and typhoid}) = 0.2 \times 0.1 $$
$$ = 0.02 $$
Answer: 0.02 (2%)
Example 2: In a health screening, the probability of detecting high blood pressure is 0.3, and the probability of detecting high blood sugar is 0.25. What is the probability that a patient has both conditions?
Given:
\( P(\text{high blood pressure}) = 0.3 \)
\( P(\text{high blood sugar}) = 0.25 \)
$$ P(\text{both}) = 0.3 \times 0.25 $$
$$ = 0.075 $$
Answer: 0.075 (7.5%)
Example 3: The probability that a patient receives vaccination A is 0.7, and vaccination B is 0.8. What is the probability that a patient receives both vaccines?
Given:
\( P(A) = 0.7 \)
\( P(B) = 0.8 \)
$$ P(A \text{ and } B) = 0.7 \times 0.8 $$
$$ = 0.56 $$
Answer: 0.56 (56%)
Example 4: If the probability of a health worker attending a training is 0.6 and the probability of the same worker receiving a promotion is 0.2, what is the probability that both happen?
Given:
\( P(\text{training}) = 0.6 \)
\( P(\text{promotion}) = 0.2 \)
$$ P(\text{training and promotion}) = 0.6 \times 0.2 $$
$$ = 0.12 $$
Answer: 0.12 (12%)
Example 5: The probability that a randomly selected household uses mosquito nets is 0.9, and the probability that the household has a child under five is 0.3. Assuming independence, what is the probability the household uses mosquito nets and has a child under five?
Given:
\( P(\text{mosquito nets}) = 0.9 \)
\( P(\text{child under five}) = 0.3 \)
$$ P(\text{both}) = 0.9 \times 0.3 $$
$$ = 0.27 $$
Answer: 0.27 (27%)
Mutually exclusive events cannot occur at the same time. The probability of either event \( A \) or event \( B \) occurring is:
$$ P(A \text{ or } B) = P(A) + P(B) $$
Example 1: In a health camp, the probability that a patient has malaria is 0.3, and the probability that the patient has typhoid is 0.2. What is the probability that the patient has either malaria or typhoid, assuming the conditions are mutually exclusive?
Given:
\( P(\text{malaria}) = 0.3 \)
\( P(\text{typhoid}) = 0.2 \)
$$ P(\text{malaria or typhoid}) = 0.3 + 0.2 $$
$$ = 0.5 $$
Answer: 0.5 (50%)
Example 2: A patient can be diagnosed with either pneumonia or bronchitis, but not both. The probabilities are 0.4 and 0.1 respectively. What is the probability of having either disease?
Given:
\( P(\text{pneumonia}) = 0.4 \)
\( P(\text{bronchitis}) = 0.1 \)
$$ P(\text{either}) = 0.4 + 0.1 $$
$$ = 0.5 $$
Answer: 0.5 (50%)
Example 3: In a clinic, the probability of a patient having flu is 0.25, and the probability of having a cold is 0.35. These are mutually exclusive. Find the probability of having flu or cold.
Given:
\( P(\text{flu}) = 0.25 \)
\( P(\text{cold}) = 0.35 \)
$$ P(\text{flu or cold}) = 0.25 + 0.35 $$
$$ = 0.6 $$
Answer: 0.6 (60%)
Example 4: The probability of a patient being diagnosed with diabetes is 0.15, and the probability of being diagnosed with hypertension is 0.25. These are mutually exclusive events. Calculate the probability of either diagnosis.
Given:
\( P(\text{diabetes}) = 0.15 \)
\( P(\text{hypertension}) = 0.25 \)
$$ P(\text{diabetes or hypertension}) = 0.15 + 0.25 $$
$$ = 0.4 $$
Answer: 0.4 (40%)
Example 5: At a health facility, the probability that a patient is tested positive for HIV is 0.1, and the probability that the patient is tested positive for hepatitis B is 0.05. These are mutually exclusive. What is the probability that the patient tests positive for either disease?
Given:
\( P(\text{HIV}) = 0.1 \)
\( P(\text{hepatitis B}) = 0.05 \)
$$ P(\text{HIV or hepatitis B}) = 0.1 + 0.05 $$
$$ = 0.15 $$
Answer: 0.15 (15%)
When two events can occur simultaneously, the probability of either event \( A \) or \( B \) occurring is:
$$ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) $$
This accounts for the overlap where both events happen.
Example 1: In a community, the probability that a person has malaria is 0.3, the probability of having typhoid is 0.2, and the probability of having both is 0.05. Calculate the probability that a person has malaria or typhoid.
Given:
\( P(\text{malaria}) = 0.3 \)
\( P(\text{typhoid}) = 0.2 \)
\( P(\text{both}) = 0.05 \)
$$ P(\text{malaria or typhoid}) = 0.3 + 0.2 - 0.05 $$
$$ = 0.45 $$
Answer: 0.45 (45%)
Example 2: The probability that a patient has high blood pressure is 0.4, the probability of diabetes is 0.3, and the probability of both conditions is 0.1. Find the probability the patient has high blood pressure or diabetes.
Given:
\( P(\text{hypertension}) = 0.4 \)
\( P(\text{diabetes}) = 0.3 \)
\( P(\text{both}) = 0.1 \)
$$ P(\text{hypertension or diabetes}) = 0.4 + 0.3 - 0.1 $$
$$ = 0.6 $$
Answer: 0.6 (60%)
Example 3: In a screening, the probability of detecting anemia is 0.25, the probability of detecting malaria is 0.35, and the probability of detecting both is 0.1. Calculate the probability of detecting anemia or malaria.
Given:
\( P(\text{anemia}) = 0.25 \)
\( P(\text{malaria}) = 0.35 \)
\( P(\text{both}) = 0.1 \)
$$ P(\text{anemia or malaria}) = 0.25 + 0.35 - 0.1 $$
$$ = 0.5 $$
Answer: 0.5 (50%)
Example 4: The probability of a patient having tuberculosis is 0.15, the probability of HIV is 0.2, and the probability of both is 0.05. What is the probability that a patient has tuberculosis or HIV?
Given:
\( P(\text{tuberculosis}) = 0.15 \)
\( P(\text{HIV}) = 0.2 \)
\( P(\text{both}) = 0.05 \)
$$ P(\text{tuberculosis or HIV}) = 0.15 + 0.2 - 0.05 $$
$$ = 0.3 $$
Answer: 0.3 (30%)
Example 5: In a health survey, the probability that a household uses treated water is 0.7, the probability that the household uses mosquito nets is 0.6, and the probability that the household uses both is 0.5. Calculate the probability that a household uses treated water or mosquito nets.
Given:
\( P(\text{treated water}) = 0.7 \)
\( P(\text{mosquito nets}) = 0.6 \)
\( P(\text{both}) = 0.5 \)
$$ P(\text{treated water or mosquito nets}) = 0.7 + 0.6 - 0.5 $$
$$ = 0.8 $$
Answer: 0.8 (80%)
In a clinic, 40% of patients have hypertension, and 25% have diabetes. If 10% have both, what is the probability that a patient has either hypertension or diabetes? (5 marks)
The probability that a community member uses a latrine is 0.85. What is the probability that a randomly selected member does not use a latrine? (3 marks)
In a survey, the probability of a household owning a bed net is 0.6, and owning a water filter is 0.4. Assuming independence, what is the probability that a household owns both? (5 marks)
Among 150 patients, 45 have malaria and 30 have typhoid. Assuming these are mutually exclusive, what is the probability that a patient has either disease? (4 marks)
The probability that a patient receives treatment A is 0.5, and treatment B is 0.7. If these events are independent, what is the probability that the patient receives neither treatment? (6 marks)
Determining the population mean and population proportion is essential for community health professionals to make evidence-based decisions. In Kenya, health workers often analyze data such as average blood pressure levels or the proportion of patients vaccinated in a county. These statistics help inform policy, resource allocation, and intervention strategies at health facilities and community programs.
The population mean is the average value of a quantitative variable for the entire population. It summarizes central tendency, providing a single value that best represents the data set. The formula to calculate the population mean is:
$$ \mu = \frac{\Sigma X_i}{N} $$
where \(\mu\) is the population mean, \(X_i\) are the individual values, and \(N\) is the population size.
Example 1: A community health survey measured the systolic blood pressure (mmHg) of 5 patients: 120, 130, 125, 135, and 140. Find the population mean systolic blood pressure.
Given: \(X_1=120\), \(X_2=130\), \(X_3=125\), \(X_4=135\), \(X_5=140\), \(N=5\)
$$ \mu = \frac{120 + 130 + 125 + 135 + 140}{5} $$
$$ \mu = \frac{650}{5} $$
$$ \mu = 130 \text{ mmHg} $$
Answer: 130 mmHg
Example 2: A sample of 8 patients' fasting blood sugar levels (mg/dL) were recorded as 90, 95, 100, 110, 105, 98, 92, and 108. Calculate the population mean blood sugar level.
Given: \(X_i = \{90,95,100,110,105,98,92,108\}\), \(N=8\)
$$ \mu = \frac{90 + 95 + 100 + 110 + 105 + 98 + 92 + 108}{8} $$
$$ \mu = \frac{798}{8} $$
$$ \mu = 99.75 \text{ mg/dL} $$
Answer: 99.75 mg/dL
Example 3: In a health center, the waiting times (minutes) for 10 patients were recorded as 15, 20, 18, 22, 19, 17, 25, 21, 16, and 23. Compute the mean waiting time.
Given: \(X_i = \{15,20,18,22,19,17,25,21,16,23\}\), \(N=10\)
$$ \mu = \frac{15 + 20 + 18 + 22 + 19 + 17 + 25 + 21 + 16 + 23}{10} $$
$$ \mu = \frac{196}{10} $$
$$ \mu = 19.6 \text{ minutes} $$
Answer: 19.6 minutes
Example 4: A community health worker records the daily number of patients visiting a dispensary over 7 days: 45, 50, 48, 52, 47, 49, 51. Find the average daily patient count.
Given: \(X_i = \{45,50,48,52,47,49,51\}\), \(N=7\)
$$ \mu = \frac{45 + 50 + 48 + 52 + 47 + 49 + 51}{7} $$
$$ \mu = \frac{342}{7} $$
$$ \mu = 48.86 \text{ patients} $$
Answer: 48.86 patients
Example 5: The weights (kg) of 6 malnourished children were recorded as 12, 13.5, 11, 14, 12.5, and 13. Calculate the average weight.
Given: \(X_i = \{12,13.5,11,14,12.5,13\}\), \(N=6\)
$$ \mu = \frac{12 + 13.5 + 11 + 14 + 12.5 + 13}{6} $$
$$ \mu = \frac{76}{6} $$
$$ \mu = 12.67 \text{ kg} $$
Answer: 12.67 kg
Population proportion quantifies the fraction of the population exhibiting a particular characteristic, such as the proportion of children vaccinated against measles in a county. It is expressed as a decimal or percentage. The population proportion \(p\) is calculated as:
$$ p = \frac{X}{N} $$
where \(X\) is the number of individuals with the attribute and \(N\) is the total population size.
Example 1: In a sample of 200 patients, 150 have received the influenza vaccine. Calculate the proportion vaccinated.
Given: \(X=150\), \(N=200\)
$$ p = \frac{150}{200} $$
$$ p = 0.75 $$
Answer: 0.75 (75%)
Example 2: A survey in a health facility found 120 out of 400 mothers attended antenatal care (ANC) visits. Find the proportion attending ANC.
Given: \(X=120\), \(N=400\)
$$ p = \frac{120}{400} $$
$$ p = 0.30 $$
Answer: 0.30 (30%)
Example 3: At a county hospital, 85 out of 170 patients tested positive for malaria. Determine the malaria positivity proportion.
Given: \(X=85\), \(N=170\)
$$ p = \frac{85}{170} $$
$$ p = 0.5 $$
Answer: 0.5 (50%)
Example 4: In a community of 1,000 people, 620 have access to clean water. Calculate the proportion with access.
Given: \(X=620\), \(N=1000\)
$$ p = \frac{620}{1000} $$
$$ p = 0.62 $$
Answer: 0.62 (62%)
Example 5: Among 500 children surveyed, 350 were found to be underweight. Find the underweight proportion.
Given: \(X=350\), \(N=500\)
$$ p = \frac{350}{500} $$
$$ p = 0.7 $$
Answer: 0.7 (70%)
In community health, it is often impractical to study the entire population. Instead, a sample is taken, and the sample mean is used to estimate the population mean. This requires understanding the relationship between sample statistics and population parameters.
The sample mean \(\bar{x}\) is calculated as:
$$ \bar{x} = \frac{\Sigma x_i}{n} $$
where \(x_i\) are sample values and \(n\) is the sample size.
Example 1: A sample of 6 patients' pulse rates (beats per minute) are 72, 75, 70, 68, 74, and 73. Estimate the average pulse rate.
Given: \(x_i = \{72,75,70,68,74,73\}\), \(n=6\)
$$ \bar{x} = \frac{72 + 75 + 70 + 68 + 74 + 73}{6} $$
$$ \bar{x} = \frac{432}{6} $$
$$ \bar{x} = 72 \text{ bpm} $$
Answer: 72 bpm
Example 2: A sample of 10 children's heights (cm) are: 110, 112, 115, 108, 111, 113, 109, 114, 110, 112. Find the sample mean height.
Given: \(x_i = \{110,112,115,108,111,113,109,114,110,112\}\), \(n=10\)
$$ \bar{x} = \frac{110 + 112 + 115 + 108 + 111 + 113 + 109 + 114 + 110 + 112}{10} $$
$$ \bar{x} = \frac{1114}{10} $$
$$ \bar{x} = 111.4 \text{ cm} $$
Answer: 111.4 cm
Example 3: From a sample of 8 patients, the number of days hospitalized are: 5, 7, 6, 4, 8, 6, 5, 7. Calculate the mean hospital stay.
Given: \(x_i = \{5,7,6,4,8,6,5,7\}\), \(n=8\)
$$ \bar{x} = \frac{5 + 7 + 6 + 4 + 8 + 6 + 5 + 7}{8} $$
$$ \bar{x} = \frac{48}{8} $$
$$ \bar{x} = 6 \text{ days} $$
Answer: 6 days
Example 4: A sample of 12 community health workers recorded the number of home visits per week: 15, 18, 17, 20, 16, 19, 18, 17, 16, 20, 19, 18. Find the average weekly visits.
Given: \(x_i = \{15,18,17,20,16,19,18,17,16,20,19,18\}\), \(n=12\)
$$ \bar{x} = \frac{15 + 18 + 17 + 20 + 16 + 19 + 18 + 17 + 16 + 20 + 19 + 18}{12} $$
$$ \bar{x} = \frac{213}{12} $$
$$ \bar{x} = 17.75 \text{ visits} $$
Answer: 17.75 visits
Example 5: In a sample of 7 clinics, the average number of patients served daily were: 45, 50, 48, 52, 47, 49, 51. Calculate the sample mean.
Given: \(x_i = \{45,50,48,52,47,49,51\}\), \(n=7\)
$$ \bar{x} = \frac{45 + 50 + 48 + 52 + 47 + 49 + 51}{7} $$
$$ \bar{x} = \frac{342}{7} $$
$$ \bar{x} = 48.86 \text{ patients} $$
Answer: 48.86 patients
Sample proportions estimate the true population proportion when it is impossible to survey everyone. This is common in community health surveys such as estimating the proportion of households with latrines.
The sample proportion \(\hat{p}\) is:
$$ \hat{p} = \frac{x}{n} $$
where \(x\) is the number of successes in the sample and \(n\) is the sample size.
Example 1: In a survey of 150 households, 90 have access to clean water. Estimate the sample proportion.
Given: \(x=90\), \(n=150\)
$$ \hat{p} = \frac{90}{150} $$
$$ \hat{p} = 0.6 $$
Answer: 0.6 (60%)
Example 2: Out of 250 patients sampled, 175 completed their medication course. Find the sample proportion.
Given: \(x=175\), \(n=250\)
$$ \hat{p} = \frac{175}{250} $$
$$ \hat{p} = 0.7 $$
Answer: 0.7 (70%)
Example 3: A sample of 300 children shows 210 are fully immunized. Calculate the sample proportion.
Given: \(x=210\), \(n=300\)
$$ \hat{p} = \frac{210}{300} $$
$$ \hat{p} = 0.7 $$
Answer: 0.7 (70%)
Example 4: Among 400 pregnant women surveyed, 320 attended at least 4 ANC visits. Determine the sample proportion.
Given: \(x=320\), \(n=400\)
$$ \hat{p} = \frac{320}{400} $$
$$ \hat{p} = 0.8 $$
Answer: 0.8 (80%)
Example 5: In a sample of 500 patients, 375 tested HIV negative. Calculate the sample proportion.
Given: \(x=375\), \(n=500\)
$$ \hat{p} = \frac{375}{500} $$
$$ \hat{p} = 0.75 $$
Answer: 0.75 (75%)
In a community survey, the weights (kg) of 7 children were recorded as 14, 15, 13, 16, 14, 15, and 17. Calculate the population mean weight. (5 marks)
A sample of 300 patients shows 180 have hypertension. Find the population proportion of hypertensive patients. (5 marks)
The waiting times (minutes) in a clinic for 12 patients are: 10, 12, 11, 13, 14, 12, 10, 15, 11, 13, 12, 14. Find the sample mean waiting time. (5 marks)
In a survey of 250 households, 200 have access to electricity. Calculate the sample proportion with electricity access. (5 marks)
A sample of 9 community health workers recorded the number of home visits per week as: 20, 22, 24, 21, 23, 25, 22, 24, 23. Find the average number of visits. (5 marks)
This chapter focused on applying inferential statistics to draw conclusions in health care delivery. It began with the calculation of probability, explaining how to determine the likelihood of events occurring within a health context. The chapter then covered the determination of population mean and proportions, showing how to estimate average values and proportions from sample data to make informed decisions. Emphasis was placed on using these statistical tools to interpret health data accurately and support evidence-based practices. The concepts were linked to practical health care scenarios where understanding variability and uncertainty is crucial. Overall, the chapter provided a foundation for using probability and population parameters to enhance health care analysis and decision-making.
In a community health survey, the probability of a patient having malaria is 0.3. What is the probability that a randomly selected patient does not have malaria? (2 marks)
In a sample of 100 patients tested for a disease, 25 tested positive. Calculate the sample proportion of patients who tested positive. (2 marks)
The weights (in kg) of 5 newborns are 3.2, 3.5, 3.0, 3.8, and 3.6. Calculate the population mean weight of the newborns. (3 marks)
A health clinic records the following number of patients per day over a week: 45, 50, 40, 55, 60, 48, 52. Calculate the mean daily number of patients. (3 marks)
The probability that a patient will recover after treatment is 0.85. If 4 patients are treated independently, calculate the probability that exactly 3 recover. (Use binomial probability) (4 marks)
In a community, 60% of households have access to clean water. If 10 households are selected at random, find the probability that at least 8 have access to clean water. (4 marks)
A sample of 50 patients has a mean blood pressure of 130 mmHg with a population standard deviation of 15 mmHg. Calculate the standard error of the mean. (3 marks)
From a sample of 200 patients, 80 have a certain infection. Construct a 95% confidence interval for the proportion of infected patients in the population. (Use \( z = 1.96 \)) (5 marks)
The average waiting time at a clinic is 25 minutes with a standard deviation of 5 minutes. If a sample of 36 patients is taken, calculate the probability that the sample mean waiting time is less than 23 minutes. (Assume normal distribution) (5 marks)
In a survey of 120 households, 72 use mosquito nets. Test at the 5% significance level whether the proportion of households using mosquito nets is different from 60%. (Use normal approximation for testing proportions) (6 marks)
Given:
Probability of having malaria, \(P(M) = 0.3\)
Probability of not having malaria, \(P(\text{not } M) = 1 - P(M) = 1 - 0.3 = 0.7\)
Answer: \( \boxed{0.7} \)
Given:
Number tested positive = 25
Sample size \(n = 100\)
Sample proportion, \( \hat{p} = \frac{25}{100} = 0.25 \)
Answer: \( \boxed{0.25} \)
Given newborn weights: 3.2, 3.5, 3.0, 3.8, 3.6 kg
Population mean,$$\mu = \frac{\sum x_i}{N} = \frac{3.2 + 3.5 + 3.0 + 3.8 + 3.6}{5} = \frac{17.1}{5} = 3.42 \text{ kg}$$
Answer: \( \boxed{3.42 \text{ kg}} \)
Given daily patients: 45, 50, 40, 55, 60, 48, 52
Mean daily patients,$$\bar{x} = \frac{45 + 50 + 40 + 55 + 60 + 48 + 52}{7} = \frac{350}{7} = 50$$
Answer: \( \boxed{50 \text{ patients}} \)
Given:
Probability of recovery \(p = 0.85\), number of patients \(n=4\), exactly 3 recover \(k=3\)
Binomial probability formula:$$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$
Calculate:$$\binom{4}{3} = 4$$$$P(3) = 4 \times (0.85)^3 \times (0.15)^1$$$$= 4 \times 0.614125 \times 0.15$$$$= 4 \times 0.09211875 = 0.368475$$
Answer: \( \boxed{0.368} \)
Given:
\(p = 0.6\), \(n = 10\), find \(P(X \ge 8)\)
Calculate \(P(X=8) + P(X=9) + P(X=10)\) using binomial formula:$$P(X=k) = \binom{10}{k} (0.6)^k (0.4)^{10-k}$$
Calculate each term:$$P(8) = \binom{10}{8} (0.6)^8 (0.4)^2 = 45 \times 0.0168 \times 0.16 = 45 \times 0.002688 = 0.12096$$$$P(9) = \binom{10}{9} (0.6)^9 (0.4)^1 = 10 \times 0.0101 \times 0.4 = 10 \times 0.00404 = 0.0404$$$$P(10) = \binom{10}{10} (0.6)^{10} (0.4)^0 = 1 \times 0.0060 \times 1 = 0.0060$$
Sum:$$0.12096 + 0.0404 + 0.0060 = 0.16736$$ Answer: \( \boxed{0.167} \)
Given:
Population standard deviation \(\sigma = 15\) mmHg
Sample size \(n=50\)
Standard error of the mean:$$SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{50}} = \frac{15}{7.071} = 2.121 \text{ mmHg}$$
Answer: \( \boxed{2.12 \text{ mmHg}} \)
Given:
Sample size \(n=200\), infected \(x=80\)
Sample proportion,$$\hat{p} = \frac{80}{200} = 0.4$$
Confidence interval formula:$$\hat{p} \pm z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
Calculate standard error:$$SE = \sqrt{\frac{0.4 \times 0.6}{200}} = \sqrt{0.0012} = 0.03464$$
Calculate margin of error:$$ME = 1.96 \times 0.03464 = 0.0679$$
Confidence interval:$$0.4 \pm 0.0679 = (0.3321, 0.4679)$$
Answer: \( \boxed{0.33 \text{ to } 0.47} \)
Given:
Population mean \(\mu = 25\) minutes
Standard deviation \(\sigma = 5\) minutes
Sample size \(n=36\)
Calculate probability \(P(\bar{x} < 23)\)
Standard error:$$SE = \frac{5}{\sqrt{36}} = \frac{5}{6} = 0.8333$$
Calculate z-score:$$z = \frac{23 - 25}{0.8333} = \frac{-2}{0.8333} = -2.4$$
From z-tables,$$P(Z < -2.4) = 0.0082$$
Answer: \( \boxed{0.0082} \)
Given:
Sample size \(n=120\), infected \(x=72\)
Sample proportion:$$\hat{p} = \frac{72}{120} = 0.6$$
Test \(H_0: p = 0.6\) vs \(H_a: p \neq 0.6\)
Calculate standard error under null hypothesis:$$SE = \sqrt{\frac{0.6 \times 0.4}{120}} = \sqrt{0.002} = 0.0447$$
Calculate z-test statistic:$$z = \frac{\hat{p} - p_0}{SE} = \frac{0.6 - 0.6}{0.0447} = 0$$
At 5% significance, critical z-values are \(\pm 1.96\). Since \(z=0\) lies within \(-1.96\) and \(1.96\), fail to reject \(H_0\).
Answer: \( \boxed{\text{No significant difference; fail to reject } H_0} \)
Question 11 (Compulsory - 20 marks)
The Kenya Medical Research Institute (KEMRI) conducted a study on the prevalence of hypertension among adults in Nairobi. Out of a sample of 500 adults, 125 were found to have hypertension.
a) Calculate the sample proportion of adults with hypertension and interpret the result in the context of public health. (10 marks)
b) If the probability that a randomly selected adult from Nairobi has hypertension is 0.25, calculate the probability that in a group of 6 adults exactly 3 have hypertension. (10 marks)
Question 12 (20 marks)
At a county hospital, the average patient satisfaction score from a survey of 80 patients is 78 out of 100. Assuming this sample represents the population, what is the population mean satisfaction score? Additionally, if the probability of a patient recommending the hospital is 0.65, calculate the probability that out of 5 randomly selected patients, at least 4 will recommend the hospital.
Question 13 (20 marks)
A community health program in Kisumu County aims to improve immunization coverage. In a sample of 250 children, 200 were immunized.
a) Calculate the proportion of immunized children. (10 marks)
b) If the probability of a child being immunized is 0.8, find the probability that in a group of 4 children, exactly 2 are immunized. (10 marks)
Question 14 (20 marks)
A health survey in Mombasa County recorded the daily number of patients visiting a clinic over 7 days as follows: 45, 52, 48, 50, 55, 47, 53.
a) Calculate the population mean number of patients visiting the clinic per day. (10 marks)
b) If the probability that a patient waits more than 30 minutes is 0.3, find the probability that in a group of 3 patients, none waits more than 30 minutes. (10 marks)
Question 11
a) Sample proportion:$$\hat{p} = \frac{125}{500} = 0.25$$
Interpretation: 25% of adults in Nairobi sampled have hypertension, indicating a significant public health concern requiring targeted interventions.
b) Probability exactly 3 out of 6 have hypertension:
Use binomial formula:$$P(X=3) = \binom{6}{3} (0.25)^3 (0.75)^3 = 20 \times 0.015625 \times 0.421875 = 0.1328$$
Question 12
Population mean satisfaction score = 78 (assuming sample represents population).
Probability at least 4 out of 5 recommend hospital:
Calculate \(P(X \geq 4) = P(X=4) + P(X=5)\) where \(X \sim Binomial(n=5, p=0.65)\).
$$P(X=4) = \binom{5}{4} (0.65)^4 (0.35)^1 = 5 \times 0.1785 \times 0.35 = 0.3124$$$$P(X=5) = \binom{5}{5} (0.65)^5 (0.35)^0 = 1 \times 0.1160 \times 1 = 0.1160$$
$$P(X \geq 4) = 0.3124 + 0.1160 = 0.4284$$
Question 13
a) Proportion immunized:$$\hat{p} = \frac{200}{250} = 0.8$$
b) Probability exactly 2 immunized out of 4:$$P(X=2) = \binom{4}{2} (0.8)^2 (0.2)^2 = 6 \times 0.64 \times 0.04 = 0.1536$$
Question 14
a) Mean number of patients:$$\text{Mean} = \frac{45 + 52 + 48 + 50 + 55 + 47 + 53}{7} = \frac{350}{7} = 50$$ patients per day
b) Probability none waits more than 30 minutes:$$P(\text{none}) = (1 - 0.3)^3 = 0.7^3 = 0.343$$