Sure! Let’s work through each part step-by-step given that [tex]\( X \)[/tex] is a normal random variable with a mean ([tex]\(\mu\)[/tex]) of 68. We assume a standard normal distribution for simplicity since the standard deviation ([tex]\(\sigma\)[/tex]) was not provided.
### Part a) [tex]\( P(X < 68) \)[/tex]
Since [tex]\( X \)[/tex] follows a normal distribution with a mean of 68, the cumulative probability of [tex]\( X \)[/tex] being less than the mean itself is 0.5.
[tex]\[ P(X < 68) = 0.5 \][/tex]
### Part b) The probability that [tex]\( X \)[/tex] is less than 54
Given the normal distribution parameters:
[tex]\[ P(X < 54) \approx 7.7935368191928 \times 10^{-45} \][/tex]
This value is extremely close to zero.
### Part c) [tex]\( P(68 < X < 82) \)[/tex]
To find this probability, we need the cumulative probability for both [tex]\( X = 82 \)[/tex] and [tex]\( X = 68 \)[/tex]. The desired probability is the difference between these cumulative probabilities.
[tex]\[ P(68 < X < 82) \approx 0.5 \][/tex]
### Part d) [tex]\( P(X < 82) \)[/tex]
For [tex]\( X = 82 \)[/tex]:
[tex]\[ P(X < 82) \approx 1.0 \][/tex]
This means that it is almost certain that [tex]\( X \)[/tex] is less than 82.
### Part e) The probability that [tex]\( X \)[/tex] is between 54 and 82
This is the difference in cumulative probabilities between [tex]\( X = 82 \)[/tex] and [tex]\( X = 54 \)[/tex]:
[tex]\[ P(54 < X < 82) \approx 1.0 \][/tex]
### Part f) The probability [tex]\( X \)[/tex] is less than 54 or more than 68
Since the two events are mutually exclusive, the probability is the sum of the probabilities of each event:
[tex]\[ P(X < 54 \text{ or } X > 68) \approx 7.7935368191928 \times 10^{-45} + (1 - 0.5) = 0.5 \][/tex]
To summarize:
a) [tex]\( P(X < 68) = 0.5 \)[/tex]
b) [tex]\( P(X < 54) \approx 7.7935368191928 \times 10^{-45} \)[/tex]
c) [tex]\( P(68 < X < 82) = 0.5 \)[/tex]
d) [tex]\( P(X < 82) \approx 1.0 \)[/tex]
e) [tex]\( P(54 < X < 82) \approx 1.0 \)[/tex]
f) [tex]\( P(X < 54 \text{ or } X > 68) = 0.5 \)[/tex]
