The thickness, SSS mm, of a mechanical shim has a normal distribution S∼N(1.45,0.042)S \sim N(1.45, 0.04^2)S∼N(1.45,0.042) and the thickness, KKK mm, of a spacer has a normal distribution K∼N(1.52,0.052)K \sim N(1.52, 0.05^2)K∼N(1.52,0.052). An engineer uses 5 independent shims and one spacer for a high-precision assembly. Find the probability that the combined thickness of the 5 shims is less than 5 times the thickness of the spacer.
Two independent random samples X1,X2,X3,X4,X5X_1, X_2, X_3, X_4, X_5X1,X2,X3,X4,X5 and Y1,Y2,Y3,Y4,Y5Y_1, Y_2, Y_3, Y_4, Y_5Y1,Y2,Y3,Y4,Y5 are each taken from a normal population with mean μ\muμ and standard deviation σ\sigmaσ.
Find the distribution of the random variable D=Y1−XˉD = Y_1 - \bar{X}D=Y1−Xˉ.
Hence show that P(Y1>Xˉ+σ)=0.1807P(Y_1 > \bar{X} + \sigma) = 0.1807P(Y1>Xˉ+σ)=0.1807 correct to 4 decimal places.
A researcher believes that P(U1>Uˉ+σ)=0.1807P(U_1 > \bar{U} + \sigma) = 0.1807P(U1>Uˉ+σ)=0.1807 for any random sample U1,U2,U3,U4,U5U_1, U_2, U_3, U_4, U_5U1,U2,U3,U4,U5 taken from the same normal population. Explain briefly why the result from part (b) should not be used to confirm the researcher's belief.
Find, correct to 3 decimal places, the actual value of P(U1>Uˉ+σ)P(U_1 > \bar{U} + \sigma)P(U1>Uˉ+σ).
616 exam-style questions on Edexcel A Level Maths The Normal Distribution, covering 3.1 The Normal Distribution, 3.2 Finding Probabilities for Normal Distributions, 3.3 The Inverse Normal Distribution Function, 3.4 The Standard Normal Distribution, 3.5 Finding the mean and standard deviation, 3.6 Approximating a Binomial Distribution, and 3.7 Hypothesis Testing with the Normal Distribution. Each one has a worked solution and a mark scheme showing where the marks go.