Answer :
Answer:
k = 75.1
Step-by-step explanation:
To solve for k such that P(72 ≤ X ≤ k) = 0.1, given that X is normally distributed with mean μ = 80 and standard deviation σ = 10, we can use the properties of the standard normal distribution.
In the context of a normal distribution with mean μ = 80 and standard deviation σ = 10, the probability P(X ≤ 72) represents the area under the normal curve to the left of X = 72. This area under the curve represents the cumulative probability that X takes on a value less than or equal to 72.
Therefore, the probability P(72 ≤ X ≤ k) can be found by subtracting the area under the normal curve to the left of X = 72 from the area under the normal curve to the left of X = k:
[tex]\sf P(72 \leq X \leq k) = P(X \leq k) - P(X \leq 72)[/tex]
Rearrange the equation to isolate P(X ≤ k):
[tex]\sf P(X \leq k)=P(72 \leq X \leq k) + P(X \leq 72)[/tex]
Given that P(72 ≤ X ≤ k) = 0.1 and P(X ≤ 72) = 0.212, then:
[tex]\sf P(X \leq k)=0.1+0.212 \\\\P(X \leq k)=0.312[/tex]
To find k, input the following into the inverse normal function on a statistical calculator:
- Area = 0.312
- σ = 10
- μ = 80
Therefore, P(X ≤ k) = 0.312 for:
[tex]k = 75.0981082...\\\\k=75.1\; \sf (3\;s.f.)[/tex]
So, the value of k such that P(72 ≤ X ≤ k) = 0.1 is:
[tex]\LARGE\boxed{\boxed{k=75.1 }}[/tex]
