To solve this problem, we use the binomial distribution as a model. The number of members with high blood pressure is expected to follow a binomial distribution with parameters $n = 300$ (the total members sampled) and $p = 0.20$ (the probability of a member having high blood pressure).

1. Expected Number of Members with High Blood Pressure

The expected number $E(X)$ is calculated as: $E(X) = n \cdot p$ Substituting the values: $E(X) = 300 \cdot 0.20 = 60$ So, 60 members are expected to have high blood pressure.

2. Standard Deviation

The standard deviation $\sigma$ for a binomial distribution is given by: $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$ Substituting the values: $\sigma = \sqrt{300 \cdot 0.20 \cdot (1 - 0.20)} = \sqrt{300 \cdot 0.20 \cdot 0.80} = \sqrt{48} \approx 6.93$ So, the standard deviation is approximately 6.93.

Final Answer:

• Expected number: 60 members
• Standard deviation: Approximately 6.93 members

Would you like me to elaborate further or perform additional calculations?

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Here are some related questions:

1. What is the probability that exactly 50 members have high blood pressure in this sample?
2. What is the probability that fewer than 55 members have high blood pressure?
3. How does the normal approximation apply to this binomial distribution?
4. What is the z-score for 70 members having high blood pressure?
5. How would the expected value change if $p$ increased to 0.25?

Tip: When dealing with binomial probabilities, understanding the expected value and standard deviation helps determine the spread of the distribution effectively!