Case 51Bernoulli distribution models the individual failures.
Note: questions 2 and 3 relate to material in chapter 6 and should
be deleted here2For the mower test data, the sampling distribution
of the proportion is approximately normalp = 54/3000 =
0.018standard error = sqrt(0.018(1-0.018)/3000) = 0.0024273442
3yes40.0185Binomial, n = 100; p =
0.018xf(x)00.162610572410.298064185820.270443166530.161935419540.071980458950.025332430360.007352080170.00180967780.00038561690.0000722539100.0000120521110.0000018075120.0000002457130.0000000305140.0000000035150.00000000041601701801902006Average
blade weigtht = 4.99Standard deviation = 0.11Using the empirical
rules, we might expect 95% of blade weights to fall within 4.99 +/-
2*0.117P(weight < 5.20) = 0.9718748174P(weight > 5.20)
=0.02812518268P(weight < 4.80) = 0.04205934749actual number >
5.27(use countif function)actual number < 4.88Percentage =
4.29%comparing this with the answers to questions 7 and 8 shows
that the predicted percentage using the normal distribution is
somewhat larger10The process is generally stable except for a clear
spike in the middle11By computing z-scores for the observations, we
see that observation 171 has a z-score of 8.04;observation 172 has
a score of 3.1; andobservation 37 has a z-score of -3.3. Obs 171 is
clearly and outlier, and the others might be considered as outliers
also.12BinFrequency4.884.85164.9464.95645555.05715.1485.15265.29More7The
histogram appears to be approximately normalUsing Risk Solver
Platform, the best fitted distribution is an "inverse Gaussian,"
which is not described in the book.However, we see that the normal
distribution is a very close fit also.
