To solve this problem, we need to determine the mean and standard deviation of [tex]\( X \)[/tex], the number of days with an afternoon tropical storm during Abraham's week-long stay. This is a binomial distribution problem where each day is an independent trial with two possible outcomes: either it rains or it doesn't.
Given:
- The number of days in a week: [tex]\( n = 7 \)[/tex]
- The probability of rain on any given afternoon: [tex]\( p = 0.7 \)[/tex]
### Mean of [tex]\( X \)[/tex]
The mean of a binomial distribution is given by the formula:
[tex]\[ \mu_X = n \cdot p \][/tex]
Substituting in the given values:
[tex]\[ \mu_X = 7 \cdot 0.7 = 4.9 \][/tex]
### Standard Deviation of [tex]\( X \)[/tex]
The standard deviation of a binomial distribution is given by the formula:
[tex]\[ \sigma_X = \sqrt{n \cdot p \cdot (1 - p)} \][/tex]
Substituting in the given values:
[tex]\[ \sigma_X = \sqrt{7 \cdot 0.7 \cdot (1 - 0.7)} \][/tex]
[tex]\[ \sigma_X = \sqrt{7 \cdot 0.7 \cdot 0.3} \][/tex]
[tex]\[ \sigma_X \approx 1.21 \][/tex]
Based on this detailed analysis, the correct mean and standard deviation of [tex]\( X \)[/tex] are:
[tex]\[ \mu_X = 4.9 \][/tex]
[tex]\[ \sigma_X \approx 1.21 \][/tex]
These values match the first option provided, but the exact numerical option is not listed in the choices. The result of the calculation should be approximately:
[tex]\[ \mu_X \approx 4.9 \][/tex]
[tex]\[ \sigma_X \approx 1.21 \][/tex]
