Answer: 0.1
Step-by-step explanation:
Let x be a random variable that is uniformly distributed on interval [a,b]
The probability density function for x :-
[tex]f(x)=\dfrac{1}{b-a}[/tex]
Given : A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 46.0 and 56.0 minutes.
Let x be the random variable that denotes the lengths of her classes.
The probability density function = [tex]f(x)=\dfrac{1}{56-46}=\dfrac{1}{10}[/tex]
Now, the probability that a given class period runs between 50.75 and 51.75 minutes will be :-
[tex]P(50.75<x<51.75)=\int^{51.75}_{50.75}\ f(x)\ dx\\\\=\int^{51.75}_{50.75}\ \dfrac{1}{10}\ dx\\\\=\dfrac{1}{10}[x]^{51.75}_{50.75}\\\\=\dfrac{1}{10}[51.75-50.75]=\dfrac{1}{10}(1)=0.1[/tex]
Hence, the required probability = 0.1
