Certainly! Let's walk through the steps to solve the given problem using the provided results.
### Part 1 of 5
Question: Is it appropriate to use the normal approximation to find the probability that more than 43% of the people in the sample have high blood pressure?
Solution:
To determine if the normal approximation is appropriate, we need to check two conditions:
1. [tex]\( np \geq 10 \)[/tex]
2. [tex]\( n(1 - p) \geq 10 \)[/tex]
Given:
- [tex]\( p = 0.3 \)[/tex]
- [tex]\( n = 33 \)[/tex]
Calculate [tex]\( np \)[/tex]:
[tex]\[ np = 33 \times 0.3 = 9.9 \][/tex]
Calculate [tex]\( n(1 - p) \)[/tex]:
[tex]\[ n(1 - p) = 33 \times (1 - 0.3) = 33 \times 0.7 = 23.1 \][/tex]
Since [tex]\( np = 9.9 \)[/tex] which is slightly less than 10, the condition is not fully met. Hence, it:
[tex]\[ \boxed{\text{is not appropriate}} \][/tex]
It is not appropriate to use the normal curve, since [tex]\( np = 9.9 \)[/tex] is not [tex]\( \geq 10 \)[/tex] and [tex]\( n(1 - p) = 23.1 \)[/tex] is [tex]\( \geq 10 \)[/tex].
### Part 2 of 5
Question: A new sample of 80 adults is drawn. Find the probability that more than 38% of the people in this sample have high blood pressure.
Solution:
Given for the new sample:
- [tex]\( p = 0.3 \)[/tex]
- [tex]\( n_{\text{new}} = 80 \)[/tex]
- [tex]\( p_{\text{new}} = 0.38 \)[/tex]
To find this probability, we need to use the standard normal distribution (Z-score).
Mean:
[tex]\[ \mu = np = 80 \times 0.3 = 24 \][/tex]
Standard Deviation:
[tex]\[ \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{80 \times 0.3 \times 0.7} = \sqrt{16.8} \approx 4.1 \][/tex]
Z-score:
[tex]\[ Z = \frac{(p_{\text{new}} - p)}{(\sigma / \sqrt{n})} = \frac{(0.38 - 0.3)}{(4.1 / \sqrt{80})} \approx 0.4307 \][/tex]
The probability that more than 38% of the people have high blood pressure is the area to the right of 0.4307 in the standard normal distribution. Using the Z-table or calculator, this probability is:
[tex]\[ \boxed{0.4307} \][/tex]
### Part 3 of 5
Question: Find the probability that the proportion of individuals in the sample of 80 who have high blood pressure is between 0.24 and 0.35.
Solution:
Given:
- [tex]\( n_{\text{new}} = 80 \)[/tex]
- [tex]\( p = 0.3 \)[/tex]
- Lower bound [tex]\( p_L = 0.24 \)[/tex]
- Upper bound [tex]\( p_U = 0.35 \)[/tex]
For the lower bound:
[tex]\[ Z_L = \frac{(p_L - p)}{(\sigma / \sqrt{n})} = \frac{(0.24 - 0.3)}{(4.1 / \sqrt{80})} \approx -0.5204 \][/tex]
For the upper bound:
[tex]\[ Z_U = \frac{(p_U - p)}{(\sigma / \sqrt{n})} = \frac{(0.35 - 0.3)}{(4.1 / \sqrt{80})} \approx 1.3012 \][/tex]
The probability that the proportion of individuals with high blood pressure is between 0.24 and 0.35 is the area between [tex]\( Z = -0.5204 \)[/tex] and [tex]\( Z = 1.3012 \)[/tex].
Using the Z-table or calculator, this probability is approximately:
[tex]\[ \boxed{0.0955} \][/tex]
In summary:
- [tex]\( n(1-p) = 23.1 \)[/tex]
- It is not appropriate to use the normal approximation.
- The probability that more than 38% of the sample (of 80 people) have high blood pressure is 0.4307.
- The probability that the proportion is between 0.24 and 0.35 is 0.0955.
