🧠:<think>First, I need to calculate the area of the region between the two circles. The larger circle has a radius that is twice the radius of the smaller circle. Given that the smaller circle has a radius of 3 units, the larger circle must have a radius of 6 units.To find the area between the two circles, I'll calculate the area of the larger circle and subtract the area of the smaller circle. The formula for the area of a circle is ( pi r^2 ).For the larger circle:( text{Area}_{text{large}} = pi times 6^2 = 36pi )For the smaller circle:( text{Area}_{text{small}} = pi times 3^2 = 9pi )Subtracting the two areas gives:( text{Area}_{text{between}} = 36pi - 9pi = 27pi )Next, I need to determine the length of the diagonal of a square inscribed in the larger circle. The diagonal of the square is equal to the diameter of the circle. Since the radius of the larger circle is 6 units, the diameter is:( text{Diameter} = 2 times 6 = 12 ) unitsTherefore, the diagonal of the square is 12 units.</think>

🧠:<think>Alright, so I have these two probability questions to solve. Both involve comparing two normal distributions, which I remember is a common thing in statistics. Let me take them one by one.Starting with the first problem: It's about the number of units sold for packaging types A and B. The question is asking for the probability that packaging B outperforms A by at least 30 units on a randomly selected day.Okay, so both A and B follow normal distributions. For A, the mean is 500 units with a standard deviation of 50. For B, the mean is 530 units with a standard deviation of 45. I need to find the probability that B - A is at least 30.Hmm, right. When dealing with the difference of two independent normal variables, the result is also a normal distribution. The mean of the difference will be the difference of the means, and the variance will be the sum of the variances.So, let me define D = B - A. Then, the mean of D, μ_D, is μ_B - μ_A = 530 - 500 = 30 units. The variance of D, σ_D², is σ_B² + σ_A² = 45² + 50². Let me calculate that: 45 squared is 2025, and 50 squared is 2500, so total variance is 2025 + 2500 = 4525. Therefore, the standard deviation σ_D is the square root of 4525.Let me compute that. The square root of 4525. Hmm, 67 squared is 4489, which is close. 67*67=4489, so 67.2 squared is 67.2*67.2. Let me compute 67*67=4489, 67*0.2=13.4, 0.2*67=13.4, and 0.2*0.2=0.04. So, (67 + 0.2)^2 = 67² + 2*67*0.2 + 0.2² = 4489 + 26.8 + 0.04 = 4515.84. Hmm, that's still less than 4525. So, 67.2² = 4515.84, so 4525 - 4515.84 = 9.16. So, we need an additional 9.16/(2*67.2) ≈ 9.16/134.4 ≈ 0.068. So, approximately 67.2 + 0.068 ≈ 67.268. So, σ_D ≈ 67.27.So, D ~ N(30, 67.27²). Now, we need P(D ≥ 30). Wait, that's interesting because the mean of D is 30. So, we're looking for the probability that D is at least its mean. In a normal distribution, the probability of being above the mean is 0.5. So, is it 0.5?Wait, hold on. Let me think again. The question is asking for the probability that B outperforms A by at least 30 units. So, D = B - A ≥ 30. Since D has a mean of 30, we're looking for P(D ≥ 30). Since the distribution is symmetric around the mean, the probability that D is greater than or equal to 30 is 0.5.But wait, that seems too straightforward. Let me double-check. Maybe I made a mistake in calculating the variance or something.Wait, no. The mean difference is 30, so the distribution of D is centered at 30. So, the probability that D is greater than or equal to 30 is indeed 0.5. Hmm, that seems right. So, the answer is 0.5 or 50%.But wait, let me confirm. Maybe I should standardize it. Let's compute Z = (D - μ_D)/σ_D. So, Z = (30 - 30)/67.27 = 0. So, P(D ≥ 30) = P(Z ≥ 0) = 0.5. Yep, that's correct.Okay, so the first probability is 0.5.Moving on to the second problem: It's about average customer ratings. Packaging A has an average rating of 7.5 with a standard deviation of 1.2, and Packaging B has an average rating of 8.2 with a standard deviation of 1.0. We need to find the probability that the average rating for B is higher than A by at least 0.5 points.Again, this is similar to the first problem. We can define the difference in ratings, say, R = B - A. Then, R will follow a normal distribution with mean μ_R = μ_B - μ_A = 8.2 - 7.5 = 0.7. The variance of R is σ_B² + σ_A² = 1.0² + 1.2² = 1 + 1.44 = 2.44. So, the standard deviation σ_R is sqrt(2.44). Let me compute that.sqrt(2.44). Well, 1.5² is 2.25, and 1.56² is 2.4336, which is very close to 2.44. So, sqrt(2.44) ≈ 1.56.So, R ~ N(0.7, 1.56²). We need to find P(R ≥ 0.5). So, we can standardize this.Z = (R - μ_R)/σ_R = (0.5 - 0.7)/1.56 = (-0.2)/1.56 ≈ -0.1282.So, we need P(Z ≥ -0.1282). Since the normal distribution is symmetric, P(Z ≥ -0.1282) = P(Z ≤ 0.1282). Looking up 0.1282 in the standard normal table.Wait, let me recall the Z-table values. For Z = 0.13, the cumulative probability is approximately 0.5517. Since 0.1282 is slightly less than 0.13, maybe around 0.5505 or something. But let me be precise.Alternatively, using a calculator or precise Z-table. Let me recall that for Z = 0.12, the cumulative is about 0.5478, and for Z = 0.13, it's 0.5517. So, 0.1282 is 0.12 + 0.0082. The difference between 0.12 and 0.13 is 0.0039 in cumulative probability. So, 0.0082 is roughly 82% of the interval from 0.12 to 0.13. So, 0.5478 + 0.0039*(0.82) ≈ 0.5478 + 0.0032 ≈ 0.5510.So, approximately 0.5510. Therefore, P(R ≥ 0.5) ≈ 0.5510.Wait, but let me think again. The question is about the probability that the average rating for B is higher than A by at least 0.5 points. So, R = B - A ≥ 0.5. So, we have R ~ N(0.7, 1.56²). So, the probability that R is at least 0.5 is the same as 1 - P(R < 0.5). But since we're dealing with a normal distribution, we can compute the Z-score as (0.5 - 0.7)/1.56 ≈ -0.1282, and then find the area to the right of that Z-score, which is 1 - Φ(-0.1282) = Φ(0.1282) ≈ 0.5510.So, yes, approximately 55.1%.But let me verify my calculations again. The mean difference is 0.7, which is already higher than 0.5, so the probability should be more than 0.5, which aligns with 0.5510.Alternatively, if I use a more precise method, maybe using a calculator or software, but since I don't have that, my approximation is 0.5510.So, summarizing:1. The probability that B outperforms A by at least 30 units is 0.5.2. The probability that B's average rating is higher than A's by at least 0.5 points is approximately 0.551.Wait, but let me think about the second problem again. The question says "the average rating for packaging type B is higher than the average rating for packaging type A by at least 0.5 points." So, is that R ≥ 0.5? Yes, because R = B - A, so R ≥ 0.5.But the mean of R is 0.7, so 0.5 is below the mean. So, the probability should be more than 0.5, which matches our result of ~0.551.Alternatively, if the question had been "at least 0.7 points," then the probability would have been 0.5. But since it's 0.5, which is less than the mean difference, the probability is higher than 0.5.So, I think my calculations are correct.Final Answer1. The probability is boxed{0.5}.2. The probability is approximately boxed{0.551}.</think>

🧠:<think>Okay, so I have this problem about a gymnast's curvature function, which is modeled by ( f(x) = a sin(bx) + c ). The interval is from (-2) to (2). The first part asks me to find the constants ( a ), ( b ), and ( c ) given that the maximum curvature is 3 units, the minimum is -1 unit, and the period is ( frac{4pi}{3} ).Alright, let's break this down. I remember that for a sine function of the form ( f(x) = a sin(bx) + c ), the amplitude is ( |a| ), the period is ( frac{2pi}{b} ), and the vertical shift is ( c ). So, the maximum value of the function is ( c + a ) and the minimum is ( c - a ).Given that the maximum curvature is 3 and the minimum is -1, I can set up two equations:1. ( c + a = 3 )2. ( c - a = -1 )If I add these two equations together, I get:( (c + a) + (c - a) = 3 + (-1) )( 2c = 2 )( c = 1 )Now, plugging ( c = 1 ) back into the first equation:( 1 + a = 3 )( a = 2 )So, ( a = 2 ) and ( c = 1 ).Next, the period is given as ( frac{4pi}{3} ). The period of a sine function is ( frac{2pi}{b} ), so:( frac{2pi}{b} = frac{4pi}{3} )Solving for ( b ):Multiply both sides by ( b ):( 2pi = frac{4pi}{3} b )Divide both sides by ( frac{4pi}{3} ):( b = frac{2pi}{frac{4pi}{3}} = frac{2pi times 3}{4pi} = frac{6pi}{4pi} = frac{3}{2} )So, ( b = frac{3}{2} ).Let me double-check these calculations. The amplitude is 2, which makes the maximum 3 and minimum -1 when added to the vertical shift of 1. The period calculation seems right: ( frac{2pi}{b} = frac{4pi}{3} ) leads to ( b = frac{3}{2} ). Yeah, that seems correct.So, the function is ( f(x) = 2 sinleft(frac{3}{2}xright) + 1 ).Moving on to the second part. I need to compute the second derivative ( f''(x) ) and find the points where the curvature is changing most rapidly on the interval ([-2, 2]).First, let's find the first derivative ( f'(x) ). The derivative of ( sin(bx) ) is ( b cos(bx) ), so:( f'(x) = 2 times frac{3}{2} cosleft(frac{3}{2}xright) = 3 cosleft(frac{3}{2}xright) )Now, the second derivative ( f''(x) ) is the derivative of ( f'(x) ). The derivative of ( cos(bx) ) is ( -b sin(bx) ), so:( f''(x) = 3 times left(-frac{3}{2}right) sinleft(frac{3}{2}xright) = -frac{9}{2} sinleft(frac{3}{2}xright) )So, ( f''(x) = -frac{9}{2} sinleft(frac{3}{2}xright) ).The question is asking for the points where the curvature is changing most rapidly. Since ( f''(x) ) represents the rate of change of curvature, the most rapid change occurs where ( |f''(x)| ) is maximized.The function ( f''(x) ) is a sine function with amplitude ( frac{9}{2} ). The maximum absolute value of ( sin ) is 1, so the maximum of ( |f''(x)| ) is ( frac{9}{2} ).To find the points where this occurs, we need to solve ( sinleft(frac{3}{2}xright) = pm 1 ).So, ( frac{3}{2}x = frac{pi}{2} + kpi ), where ( k ) is an integer.Solving for ( x ):( x = frac{pi}{2} times frac{2}{3} + kpi times frac{2}{3} )( x = frac{pi}{3} + frac{2kpi}{3} )Now, we need to find all ( x ) in the interval ([-2, 2]) that satisfy this equation.Let's compute the possible values of ( k ) such that ( x ) is within ([-2, 2]).First, let's compute ( frac{pi}{3} approx 1.047 ), and ( frac{2pi}{3} approx 2.094 ).So, starting with ( k = 0 ):( x = frac{pi}{3} approx 1.047 ) which is within [-2, 2].( k = 1 ):( x = frac{pi}{3} + frac{2pi}{3} = pi approx 3.1416 ), which is outside the interval.( k = -1 ):( x = frac{pi}{3} - frac{2pi}{3} = -frac{pi}{3} approx -1.047 ), which is within [-2, 2].( k = -2 ):( x = frac{pi}{3} - frac{4pi}{3} = -pi approx -3.1416 ), which is outside the interval.So, the critical points within [-2, 2] are ( x = frac{pi}{3} ) and ( x = -frac{pi}{3} ).But wait, let me check if these are the only solutions. The general solution is ( x = frac{pi}{3} + frac{2kpi}{3} ). Let's plug in ( k = 0, 1, -1, 2, -2 ) and see.For ( k = 0 ): ( x = frac{pi}{3} approx 1.047 ) (inside)For ( k = 1 ): ( x = frac{pi}{3} + frac{2pi}{3} = pi approx 3.1416 ) (outside)For ( k = -1 ): ( x = frac{pi}{3} - frac{2pi}{3} = -frac{pi}{3} approx -1.047 ) (inside)For ( k = 2 ): ( x = frac{pi}{3} + frac{4pi}{3} = frac{5pi}{3} approx 5.235 ) (outside)For ( k = -2 ): ( x = frac{pi}{3} - frac{4pi}{3} = -pi approx -3.1416 ) (outside)So, only ( k = 0 ) and ( k = -1 ) give solutions within [-2, 2]. Therefore, the points where the curvature is changing most rapidly are at ( x = frac{pi}{3} ) and ( x = -frac{pi}{3} ).But wait, let me make sure. The function ( f''(x) = -frac{9}{2} sinleft(frac{3}{2}xright) ). The maximum absolute value occurs when ( sinleft(frac{3}{2}xright) = pm 1 ), which is indeed when ( frac{3}{2}x = frac{pi}{2} + kpi ), so the solutions are ( x = frac{pi}{3} + frac{2kpi}{3} ).So, within [-2, 2], the only solutions are ( x = frac{pi}{3} ) and ( x = -frac{pi}{3} ). Let me compute their approximate decimal values to ensure they are within [-2, 2].( frac{pi}{3} approx 1.047 ) which is less than 2, so it's inside.( -frac{pi}{3} approx -1.047 ), which is greater than -2, so it's inside.Therefore, these are the two points where the curvature is changing most rapidly.Wait, but let me check if there are more points. Let's see, for ( k = 1 ), ( x = frac{pi}{3} + frac{2pi}{3} = pi approx 3.1416 ), which is outside.For ( k = -1 ), ( x = -frac{pi}{3} approx -1.047 ), which is inside.Wait, but when ( k = -2 ), ( x = -frac{pi}{3} - frac{2pi}{3} = -pi approx -3.1416 ), which is outside.So, only two points: ( x = frac{pi}{3} ) and ( x = -frac{pi}{3} ).Therefore, the points where the curvature is changing most rapidly are at ( x = frac{pi}{3} ) and ( x = -frac{pi}{3} ).But let me also think about the endpoints of the interval. The interval is [-2, 2]. The function ( f''(x) ) is continuous on this interval, so the maximum and minimum of ( |f''(x)| ) must occur either at critical points or at endpoints.We found the critical points where ( f''(x) ) reaches its maximum and minimum. But we should also check the endpoints to make sure.Compute ( f''(-2) ) and ( f''(2) ):First, ( f''(x) = -frac{9}{2} sinleft(frac{3}{2}xright) ).So, ( f''(-2) = -frac{9}{2} sinleft(frac{3}{2} times (-2)right) = -frac{9}{2} sin(-3) ).Since ( sin(-3) = -sin(3) ), so:( f''(-2) = -frac{9}{2} times (-sin(3)) = frac{9}{2} sin(3) ).Similarly, ( f''(2) = -frac{9}{2} sinleft(frac{3}{2} times 2right) = -frac{9}{2} sin(3) ).Now, ( sin(3) ) is approximately ( sin(3) approx 0.1411 ). So, ( |f''(-2)| = frac{9}{2} times 0.1411 approx 0.635 ), and ( |f''(2)| = frac{9}{2} times 0.1411 approx 0.635 ).Comparing this to the maximum ( |f''(x)| ) at the critical points, which is ( frac{9}{2} approx 4.5 ). So, the endpoints have much smaller absolute values. Therefore, the maximum rate of change occurs at ( x = frac{pi}{3} ) and ( x = -frac{pi}{3} ).Therefore, the points where the curvature is changing most rapidly are at ( x = frac{pi}{3} ) and ( x = -frac{pi}{3} ).Let me just recap:1. Found ( a = 2 ), ( c = 1 ), ( b = frac{3}{2} ).2. Computed ( f''(x) = -frac{9}{2} sinleft(frac{3}{2}xright) ).3. Determined that the maximum absolute value of ( f''(x) ) is ( frac{9}{2} ), occurring at ( x = frac{pi}{3} + frac{2kpi}{3} ).4. Found within [-2, 2], only ( x = frac{pi}{3} ) and ( x = -frac{pi}{3} ) satisfy this.5. Checked endpoints and found they have much smaller absolute values, so the critical points are indeed where the maximum occurs.I think that's thorough enough. I don't see any mistakes in my reasoning.Final Answer1. The values are ( a = boxed{2} ), ( b = boxed{dfrac{3}{2}} ), and ( c = boxed{1} ).2. The curvature is changing most rapidly at ( x = boxed{dfrac{pi}{3}} ) and ( x = boxed{-dfrac{pi}{3}} ).</think>

🧠:<think>Alright, so I have this problem about a stockbroker analyzing a high-risk stock, Stock X. The returns of this stock are modeled by a continuous random variable ( R ) with a given probability density function (pdf). I need to find two things: the expected return ( E[R] ) and the probability that the return exceeds a certain threshold ( T ). Let me try to break this down step by step.First, let me write down the pdf again to make sure I have it correctly:[ f_R(r) = begin{cases} frac{lambda e^{-lambda r}}{1 - e^{-lambda b}} & text{if } 0 leq r leq b, 0 & text{otherwise}.end{cases}]So, this is a truncated exponential distribution. Normally, an exponential distribution has a pdf of ( lambda e^{-lambda r} ) for ( r geq 0 ), but here it's truncated at ( r = b ). That means all the probability mass beyond ( b ) is removed, and the pdf is scaled by ( frac{1}{1 - e^{-lambda b}} ) to ensure that the total integral over the interval [0, b] is 1.Starting with the first part: finding the expected return ( E[R] ). The expected value of a continuous random variable is given by the integral of ( r ) multiplied by its pdf over the entire range. So, in this case, it should be:[ E[R] = int_{0}^{b} r cdot f_R(r) , dr ]Plugging in the given pdf:[ E[R] = int_{0}^{b} r cdot frac{lambda e^{-lambda r}}{1 - e^{-lambda b}} , dr ]I can factor out the constants ( frac{lambda}{1 - e^{-lambda b}} ) from the integral:[ E[R] = frac{lambda}{1 - e^{-lambda b}} int_{0}^{b} r e^{-lambda r} , dr ]Now, I need to compute the integral ( int_{0}^{b} r e^{-lambda r} , dr ). This looks like a standard integral that can be solved using integration by parts. Let me recall the formula for integration by parts:[ int u , dv = uv - int v , du ]Let me set ( u = r ) and ( dv = e^{-lambda r} dr ). Then, ( du = dr ) and ( v = -frac{1}{lambda} e^{-lambda r} ).Applying integration by parts:[ int r e^{-lambda r} dr = -frac{r}{lambda} e^{-lambda r} + frac{1}{lambda} int e^{-lambda r} dr ]Compute the remaining integral:[ int e^{-lambda r} dr = -frac{1}{lambda} e^{-lambda r} + C ]Putting it all together:[ int r e^{-lambda r} dr = -frac{r}{lambda} e^{-lambda r} - frac{1}{lambda^2} e^{-lambda r} + C ]Now, evaluate this from 0 to ( b ):At ( r = b ):[ -frac{b}{lambda} e^{-lambda b} - frac{1}{lambda^2} e^{-lambda b} ]At ( r = 0 ):[ -frac{0}{lambda} e^{0} - frac{1}{lambda^2} e^{0} = -0 - frac{1}{lambda^2} = -frac{1}{lambda^2} ]Subtracting the lower limit from the upper limit:[ left( -frac{b}{lambda} e^{-lambda b} - frac{1}{lambda^2} e^{-lambda b} right) - left( -frac{1}{lambda^2} right) ]Simplify:[ -frac{b}{lambda} e^{-lambda b} - frac{1}{lambda^2} e^{-lambda b} + frac{1}{lambda^2} ]Factor out ( e^{-lambda b} ) from the first two terms:[ -frac{1}{lambda^2} e^{-lambda b} ( lambda b + 1 ) + frac{1}{lambda^2} ]So, the integral ( int_{0}^{b} r e^{-lambda r} dr ) is:[ frac{1}{lambda^2} left( 1 - e^{-lambda b} ( lambda b + 1 ) right) ]Now, going back to the expected value:[ E[R] = frac{lambda}{1 - e^{-lambda b}} cdot frac{1}{lambda^2} left( 1 - e^{-lambda b} ( lambda b + 1 ) right) ]Simplify the expression:First, ( frac{lambda}{1 - e^{-lambda b}} times frac{1}{lambda^2} = frac{1}{lambda (1 - e^{-lambda b})} )So,[ E[R] = frac{1}{lambda (1 - e^{-lambda b})} left( 1 - e^{-lambda b} ( lambda b + 1 ) right) ]Let me expand the numerator:[ 1 - e^{-lambda b} ( lambda b + 1 ) = 1 - lambda b e^{-lambda b} - e^{-lambda b} ]So,[ E[R] = frac{1 - lambda b e^{-lambda b} - e^{-lambda b}}{lambda (1 - e^{-lambda b})} ]I can factor out ( -e^{-lambda b} ) from the numerator:[ 1 - e^{-lambda b}(1 + lambda b) ]Wait, that's how it is. Alternatively, let me see if I can split the fraction:[ E[R] = frac{1}{lambda (1 - e^{-lambda b})} - frac{lambda b e^{-lambda b} + e^{-lambda b}}{lambda (1 - e^{-lambda b})} ]But that might complicate things. Alternatively, let's factor ( 1 - e^{-lambda b} ) in the numerator:Wait, the numerator is ( 1 - e^{-lambda b}(1 + lambda b) ). Let me write it as:[ 1 - e^{-lambda b} - lambda b e^{-lambda b} ]So, the numerator is ( (1 - e^{-lambda b}) - lambda b e^{-lambda b} ). Therefore, we can split the fraction:[ E[R] = frac{(1 - e^{-lambda b}) - lambda b e^{-lambda b}}{lambda (1 - e^{-lambda b})} ]Which can be written as:[ E[R] = frac{1 - e^{-lambda b}}{lambda (1 - e^{-lambda b})} - frac{lambda b e^{-lambda b}}{lambda (1 - e^{-lambda b})} ]Simplify each term:First term: ( frac{1 - e^{-lambda b}}{lambda (1 - e^{-lambda b})} = frac{1}{lambda} )Second term: ( frac{lambda b e^{-lambda b}}{lambda (1 - e^{-lambda b})} = frac{b e^{-lambda b}}{1 - e^{-lambda b}} )So, putting it together:[ E[R] = frac{1}{lambda} - frac{b e^{-lambda b}}{1 - e^{-lambda b}} ]Hmm, that seems manageable. Let me see if I can write it differently. Notice that ( frac{b e^{-lambda b}}{1 - e^{-lambda b}} ) can be expressed as ( frac{b}{e^{lambda b} - 1} ), since ( 1 - e^{-lambda b} = frac{e^{lambda b} - 1}{e^{lambda b}} ). So, let's verify:[ frac{b e^{-lambda b}}{1 - e^{-lambda b}} = frac{b e^{-lambda b}}{frac{e^{lambda b} - 1}{e^{lambda b}}} = frac{b e^{-lambda b} cdot e^{lambda b}}{e^{lambda b} - 1} = frac{b}{e^{lambda b} - 1} ]Yes, that's correct. So, substituting back:[ E[R] = frac{1}{lambda} - frac{b}{e^{lambda b} - 1} ]Alternatively, I can write this as:[ E[R] = frac{1}{lambda} left( 1 - frac{lambda b}{e^{lambda b} - 1} right) ]But I think the first expression is simpler:[ E[R] = frac{1}{lambda} - frac{b}{e^{lambda b} - 1} ]Let me check if this makes sense. When ( b ) approaches infinity, the distribution becomes the standard exponential distribution, and the expected value should approach ( frac{1}{lambda} ). Let's see:As ( b to infty ), ( e^{lambda b} ) becomes very large, so ( frac{b}{e^{lambda b} - 1} ) approaches 0. Thus, ( E[R] ) approaches ( frac{1}{lambda} ), which is correct. That's a good sanity check.Another sanity check: when ( lambda ) is very large, meaning the distribution is sharply peaked near 0, the expected value should be close to 0. Let's see:If ( lambda ) is large, ( e^{-lambda b} ) is very small, so ( 1 - e^{-lambda b} approx 1 ), and ( frac{1}{lambda} ) is small. Also, ( frac{b}{e^{lambda b} - 1} approx frac{b}{e^{lambda b}} ), which is negligible. So, ( E[R] approx frac{1}{lambda} ), which is small, as expected.Wait, but if ( lambda ) is very large, the distribution is concentrated near 0, so the expected value should be near 0. But according to the formula, it's ( frac{1}{lambda} ), which is small but not exactly 0. Hmm, is that correct?Wait, actually, for a truncated exponential distribution, the expectation is not exactly 0 even if ( lambda ) is very large because there's still a non-zero probability near 0. So, maybe it's correct. Alternatively, let me think about the case when ( b ) is very small. If ( b ) approaches 0, the distribution becomes a point mass at 0, so the expectation should approach 0. Let's see:As ( b to 0 ), ( e^{lambda b} approx 1 + lambda b + frac{(lambda b)^2}{2} ). So, ( e^{lambda b} - 1 approx lambda b + frac{(lambda b)^2}{2} ). Thus, ( frac{b}{e^{lambda b} - 1} approx frac{b}{lambda b} = frac{1}{lambda} ). So, ( E[R] = frac{1}{lambda} - frac{1}{lambda} = 0 ). That makes sense. So, as ( b to 0 ), ( E[R] to 0 ), which is correct.Okay, so the formula seems to hold up under these sanity checks. Therefore, I think the expected return is:[ E[R] = frac{1}{lambda} - frac{b}{e^{lambda b} - 1} ]Moving on to the second part: finding the probability that the return ( R ) exceeds a threshold ( T ), i.e., ( P(R > T) ).Given the pdf, the probability ( P(R > T) ) is the integral of the pdf from ( T ) to ( b ), since beyond ( b ), the pdf is 0.So,[ P(R > T) = int_{T}^{b} f_R(r) , dr ]Again, plugging in the pdf:[ P(R > T) = int_{T}^{b} frac{lambda e^{-lambda r}}{1 - e^{-lambda b}} , dr ]Factor out the constants ( frac{lambda}{1 - e^{-lambda b}} ):[ P(R > T) = frac{lambda}{1 - e^{-lambda b}} int_{T}^{b} e^{-lambda r} , dr ]Compute the integral ( int_{T}^{b} e^{-lambda r} dr ). The antiderivative of ( e^{-lambda r} ) is ( -frac{1}{lambda} e^{-lambda r} ). So,[ int_{T}^{b} e^{-lambda r} dr = left[ -frac{1}{lambda} e^{-lambda r} right]_{T}^{b} = -frac{1}{lambda} e^{-lambda b} + frac{1}{lambda} e^{-lambda T} ]Simplify:[ frac{1}{lambda} ( e^{-lambda T} - e^{-lambda b} ) ]So, plugging back into the probability expression:[ P(R > T) = frac{lambda}{1 - e^{-lambda b}} cdot frac{1}{lambda} ( e^{-lambda T} - e^{-lambda b} ) ]Simplify the constants:The ( lambda ) cancels out:[ P(R > T) = frac{ e^{-lambda T} - e^{-lambda b} }{1 - e^{-lambda b}} ]We can factor out ( e^{-lambda T} ) from the numerator, but it might not be necessary. Alternatively, let's write it as:[ P(R > T) = frac{ e^{-lambda T} - e^{-lambda b} }{1 - e^{-lambda b}} ]Alternatively, we can factor the negative sign:[ P(R > T) = frac{ e^{-lambda T} - e^{-lambda b} }{1 - e^{-lambda b}} = frac{ e^{-lambda T} (1 - e^{-lambda (b - T)} ) }{1 - e^{-lambda b}} ]But I think the first form is simpler:[ P(R > T) = frac{ e^{-lambda T} - e^{-lambda b} }{1 - e^{-lambda b}} ]Let me verify this result with some sanity checks.First, when ( T = 0 ), the probability should be 1, since all returns are between 0 and ( b ). Plugging ( T = 0 ):[ P(R > 0) = frac{ e^{0} - e^{-lambda b} }{1 - e^{-lambda b}} = frac{1 - e^{-lambda b}}{1 - e^{-lambda b}} = 1 ]That's correct.Next, when ( T = b ), the probability should be 0, since the maximum return is ( b ). Plugging ( T = b ):[ P(R > b) = frac{ e^{-lambda b} - e^{-lambda b} }{1 - e^{-lambda b}} = 0 ]Correct.Another check: when ( T ) is very large, say ( T > b ), the probability should be 0, which is consistent with the formula since ( e^{-lambda T} ) would be negligible, but in reality, since ( T ) is beyond ( b ), the integral is 0.Wait, actually, in the formula, if ( T > b ), the integral from ( T ) to ( b ) is 0, so ( P(R > T) = 0 ). So, the formula correctly gives 0 in that case.Another check: when ( lambda ) is very large, the distribution is concentrated near 0, so for ( T ) near 0, the probability should be small. Let's see:If ( lambda ) is large, ( e^{-lambda T} ) is very small if ( T > 0 ). So, ( P(R > T) approx frac{ - e^{-lambda b} }{1 - e^{-lambda b}} approx frac{ -0 }{1 - 0 } = 0 ). That makes sense.Alternatively, if ( lambda ) is very small, meaning the distribution is spread out towards ( b ), then ( e^{-lambda T} ) is significant. For example, if ( lambda to 0 ), the pdf becomes uniform over [0, b], because the exponential distribution becomes flat as ( lambda to 0 ). Let's see:As ( lambda to 0 ), ( e^{-lambda r} approx 1 - lambda r ). So, the pdf becomes approximately ( frac{lambda (1 - lambda r)}{1 - (1 - lambda b)} = frac{lambda (1 - lambda r)}{lambda b} = frac{1 - lambda r}{b} approx frac{1}{b} ), which is uniform. So, the probability ( P(R > T) ) should be ( frac{b - T}{b} ).Let's see what our formula gives:[ P(R > T) = frac{ e^{-lambda T} - e^{-lambda b} }{1 - e^{-lambda b}} ]As ( lambda to 0 ), ( e^{-lambda T} approx 1 - lambda T ) and ( e^{-lambda b} approx 1 - lambda b ). So,Numerator: ( (1 - lambda T) - (1 - lambda b) = lambda (b - T) )Denominator: ( 1 - (1 - lambda b) = lambda b )Thus,[ P(R > T) approx frac{lambda (b - T)}{lambda b} = frac{b - T}{b} ]Which is exactly the probability for a uniform distribution on [0, b]. So, that's a good check.Therefore, I'm confident that the probability ( P(R > T) ) is:[ P(R > T) = frac{ e^{-lambda T} - e^{-lambda b} }{1 - e^{-lambda b}} ]Alternatively, we can factor out ( e^{-lambda T} ) from the numerator:[ P(R > T) = frac{ e^{-lambda T} (1 - e^{-lambda (b - T)} ) }{1 - e^{-lambda b}} ]But I think the original form is simpler.So, summarizing my results:1. The expected return ( E[R] ) is ( frac{1}{lambda} - frac{b}{e^{lambda b} - 1} ).2. The probability that the return exceeds ( T ) is ( frac{ e^{-lambda T} - e^{-lambda b} }{1 - e^{-lambda b}} ).I think I've covered all the steps, checked the results with various sanity checks, and everything seems consistent. I don't see any mistakes in my calculations, so I'm ready to present the final answers.Final Answer1. The expected return is (boxed{dfrac{1}{lambda} - dfrac{b}{e^{lambda b} - 1}}).2. The probability that the return exceeds (T) is (boxed{dfrac{e^{-lambda T} - e^{-lambda b}}{1 - e^{-lambda b}}}).</think>