Answer :
To tackle these problems, we need to use the concepts of normal distribution for Scenario 1 and binomial probability for Scenario 2.
Scenario 1: Normal Distribution
Given:
- Mean ([tex]\mu[/tex]) = 98.25 °F
- Standard Deviation ([tex]\sigma[/tex]) = 0.73 °F
We'll use the Z-score formula to find probabilities for normal distribution:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
where [tex]X[/tex] is the body temperature.
Q1a. What is the probability that a randomly selected body temperature will be less than 97.0 °F?
Calculate the Z-score for 97.0 °F:
[tex]Z = \frac{97.0 - 98.25}{0.73} = \frac{-1.25}{0.73} \approx -1.71[/tex]Using the standard normal distribution table, find the probability corresponding to [tex]Z = -1.71[/tex], which is approximately 0.0436.
Therefore, the probability is approximately 0.0436 or 4.36%.
Q1b. What is the probability that a randomly selected body temperature will be greater than 99.6 °F?
Calculate the Z-score for 99.6 °F:
[tex]Z = \frac{99.6 - 98.25}{0.73} = \frac{1.35}{0.73} \approx 1.85[/tex]Using the standard normal distribution table, find the probability corresponding to [tex]Z = 1.85[/tex], which is approximately 0.9678.
The probability of being greater than 99.6 °F is [tex]1 - 0.9678 = 0.0322[/tex].
Therefore, the probability is approximately 0.0322 or 3.22%.
Q1c. What is the probability that a randomly selected body temperature will be between 98.1 °F and 99.1 °F?
Calculate the Z-score for 98.1 °F:
[tex]Z = \frac{98.1 - 98.25}{0.73} \approx -0.21[/tex]Calculate the Z-score for 99.1 °F:
[tex]Z = \frac{99.1 - 98.25}{0.73} \approx 1.15[/tex]Find the probabilities for [tex]Z = -0.21[/tex] (about 0.4168) and [tex]Z = 1.15[/tex] (about 0.8749).
The difference gives the probability:
[tex]0.8749 - 0.4168 = 0.4581[/tex]
Therefore, the probability is approximately 0.4581 or 45.81%.
Scenario 2: Binomial Probability
Given:
- Probability of being less than 18 years old [tex]p = 0.202[/tex]
- Number of residents [tex]n = 15[/tex]
Use the binomial probability formula:
[tex]P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}[/tex]
where [tex]\binom{n}{k}[/tex] is the binomial coefficient.
Q2a. What is the probability that 5 people in your group will be less than 18 years old?
Calculate using [tex]k = 5[/tex]:
[tex]P(X = 5) = \binom{15}{5} (0.202)^5 (0.798)^{10}[/tex]Solve:
[tex]P(X = 5) = 3003 \times (0.202)^5 \times (0.798)^{10} \approx 0.0818[/tex]
Therefore, the probability is approximately 0.0818 or 8.18%.
Q2b. What is the probability that at least 3 people in your group will be less than 18 years old?
We need to find [tex]P(X \geq 3)[/tex], which is [tex]1 - P(X < 3)[/tex].
Calculate [tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex].
Use the formula for each and sum them up.
Subtract from 1 to find [tex]P(X \geq 3)[/tex].
Therefore, the probability can be approximately calculated as you sum up these values.
These steps break down the probability calculations for each part of the scenarios.
