Q1 ### Understanding Continuous Random Variables and Density FunctionsThe current in a certain circuit measured by an ammeter is a continuous random variable \( X \) with the following density function:\[ f(x) = \begin{cases} 0.025x + 0.4 & \text{for } 3 \le x \le 5 \\0 & \text{otherwise} \end{cases}\]#### (a) Graph the Probability Density Function (pdf)The correct graph is shown in the fourth plot. Here's an explanation of the graph:- The function \( f(x) \) is defined for \( x \) in the interval \([3, 5]\).- For \( x \) outside this interval, the density is 0.- The pdf starts at \( (3, 0.475) \), increases linearly, reaching \( (5, 0.525) \).The graph of the pdf is a straight line starting at (3, 0.475) and ending at (5, 0.525) within the interval \([3, 5]\). Outside this interval, the graph is 0.#### Verify the Area Under the Density CurveTo verify that the total area under the density curve is indeed 1, compute the integral:\[ \int_{3}^{5} (0.025x + 0.4) \, dx \]1. Compute the indefinite integral:\[ \int (0.025x + 0.4) \, dx = 0.0125x^2 + 0.4x + C \]2. Evaluate the definite integral from 3 to 5:\[ \left[ 0.0125x^2 + 0.4x \right]_{3}^{5} = (0.0125(5)^2 + 0.4(5)) - (0.0125(3)^2 + 0.4(3))= (0.3125 + 2) - (0.1125 + 1.2)= 2.3125 - 1.3125= 1\]Thus, the area under the probability density function is indeed 1.#### (b) Calculate \( P(X \leq 4) \)To calculate \( P(X \leq 4

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The current in a certain circuit as measured by an ammeter is a continuous random variable X with the following density function. 0.025x + 0.4 3 ≤x≤5 f(x) = ) = otherwise (a) Graph the pdf. f(x) f(x) f(x) 0.6 0.5 0.4 0.3 0.2 0.1 1 2 3 4 3 O O Verify that the total area under the density curve is indeed 1. [²³ 0.025x+ 0.4 dx= X = 2.3125- ]x ) x (b) Calculate P(X ≤ 4). How does this probability compare to P(X < 4)? P(X ≤ 4) = P(X < 4) O P(X ≤ 4) > P(X < 4) O P(X ≤ 4) < P(X < 4) (c) Calculate P(3.5 ≤ x ≤ 4.5). 0.5 Calculate P(4.5 < X). X 0.6 0.5 0.4 0.3 0.2 0.1 1 2 4 5 X O 0.6 0.5 0.4 0.3 0.2 0.1 1 2 3 4 f(x) 0.6 0.5 0.4 0.3 0.2 0.1 1 2 3 4 5 X