Answer :
Sure, let's work through the problem step by step.
1. Identify the distributions:
Soledad's daily text count [tex]\( S \)[/tex] follows a normal distribution:
[tex]\( S \sim \mathcal{N}(100, 6^2) \)[/tex]
Tania's daily text count [tex]\( T \)[/tex] follows a normal distribution:
[tex]\( T \sim \mathcal{N}(108, 8.1^2) \)[/tex]
2. Define the difference in texts sent:
Let [tex]\( D = S - T \)[/tex]. Since [tex]\( S \)[/tex] and [tex]\( T \)[/tex] are independent normally distributed random variables, [tex]\( D \)[/tex] will also follow a normal distribution.
3. Calculate the mean and standard deviation of [tex]\( D \)[/tex]:
The mean of [tex]\( D \)[/tex] is given by the difference of the means of [tex]\( S \)[/tex] and [tex]\( T \)[/tex]:
[tex]\[ \text{mean}_D = \text{mean}_S - \text{mean}_T = 100 - 108 = -8 \][/tex]
The variance of [tex]\( D \)[/tex] is the sum of the variances of [tex]\( S \)[/tex] and [tex]\( T \)[/tex]:
[tex]\[ \text{var}_D = \text{var}_S + \text{var}_T = 6^2 + 8.1^2 = 36 + 65.61 = 101.61 \][/tex]
Thus, the standard deviation of [tex]\( D \)[/tex] is:
[tex]\[ \text{std}_D = \sqrt{101.61} \approx 10.08 \][/tex]
4. Find the z-score that corresponds to [tex]\( D > 0 \)[/tex]:
The z-score for [tex]\( D > 0 \)[/tex] is given by:
[tex]\[ z = \frac{0 - \text{mean}_D}{\text{std}_D} = \frac{0 - (-8)}{10.08} \approx 0.79 \][/tex]
5. Convert the z-score to a probability:
We need the cumulative probability up to [tex]\( z \approx 0.79 \)[/tex] from the standard normal distribution table. This probability is the cumulative distribution function (CDF) value.
This CDF value represents the probability that [tex]\( D \)[/tex] is less than or equal to 0. We are looking for the complementary probability [tex]\( P(D > 0) \)[/tex]:
[tex]\[ P(D > 0) = 1 - P(Z \leq 0.79) \approx 1 - 0.786 = 0.214 \][/tex]
So, the probability that Soledad sends more texts than Tania on any given day is approximately [tex]\( 0.214 \)[/tex].
Therefore, the answer is:
[tex]\[ 0.214 \][/tex]
1. Identify the distributions:
Soledad's daily text count [tex]\( S \)[/tex] follows a normal distribution:
[tex]\( S \sim \mathcal{N}(100, 6^2) \)[/tex]
Tania's daily text count [tex]\( T \)[/tex] follows a normal distribution:
[tex]\( T \sim \mathcal{N}(108, 8.1^2) \)[/tex]
2. Define the difference in texts sent:
Let [tex]\( D = S - T \)[/tex]. Since [tex]\( S \)[/tex] and [tex]\( T \)[/tex] are independent normally distributed random variables, [tex]\( D \)[/tex] will also follow a normal distribution.
3. Calculate the mean and standard deviation of [tex]\( D \)[/tex]:
The mean of [tex]\( D \)[/tex] is given by the difference of the means of [tex]\( S \)[/tex] and [tex]\( T \)[/tex]:
[tex]\[ \text{mean}_D = \text{mean}_S - \text{mean}_T = 100 - 108 = -8 \][/tex]
The variance of [tex]\( D \)[/tex] is the sum of the variances of [tex]\( S \)[/tex] and [tex]\( T \)[/tex]:
[tex]\[ \text{var}_D = \text{var}_S + \text{var}_T = 6^2 + 8.1^2 = 36 + 65.61 = 101.61 \][/tex]
Thus, the standard deviation of [tex]\( D \)[/tex] is:
[tex]\[ \text{std}_D = \sqrt{101.61} \approx 10.08 \][/tex]
4. Find the z-score that corresponds to [tex]\( D > 0 \)[/tex]:
The z-score for [tex]\( D > 0 \)[/tex] is given by:
[tex]\[ z = \frac{0 - \text{mean}_D}{\text{std}_D} = \frac{0 - (-8)}{10.08} \approx 0.79 \][/tex]
5. Convert the z-score to a probability:
We need the cumulative probability up to [tex]\( z \approx 0.79 \)[/tex] from the standard normal distribution table. This probability is the cumulative distribution function (CDF) value.
This CDF value represents the probability that [tex]\( D \)[/tex] is less than or equal to 0. We are looking for the complementary probability [tex]\( P(D > 0) \)[/tex]:
[tex]\[ P(D > 0) = 1 - P(Z \leq 0.79) \approx 1 - 0.786 = 0.214 \][/tex]
So, the probability that Soledad sends more texts than Tania on any given day is approximately [tex]\( 0.214 \)[/tex].
Therefore, the answer is:
[tex]\[ 0.214 \][/tex]
