Math 143 C/E, Spring 2001
IPS Reading Questions
Chapter 5, Section 2 (pp. 397-408)



  1. You may recall that in problems 4.52, 4.71 and 4.72 we investigated life insurance payouts. In 4.71 you found the standard deviation when there is only one observation (that is, X represented just one person insured). In 4.72(b), you found the standard deviation for the average payout Z per person with two observations. Look up the values of these two standard deviations. How do they illustrate the formula for standard deviation of a sample mean on p. 399?

    The standard deviation found in 4.71 is s, the population's standard deviation. The s.d. in 4.72(b) should be the same as that in 4.71 divided by the square root of 2 (since n = 2). You could decrease variability even more by including more observations in the calculation of the mean, illustrating the first sentence of Example 5.15.

  2. The authors indicate that, if the underlying population is normal, then so is the sampling distribution for the sample mean of n independent observations. In using the sampling distributions applet we have seen that the sampling distribution for the mean can appear normal even when the underlying population is not. Under what general conditions can you expect this sampling distribution to be normal?

    Actually, no matter what shape the underlying distribution has, the sampling distribution for the sample mean will become more and more normal for larger and larger values of n, with distribution N( m, s/n1/2). This is what the central limit theorem says. In fact, the authors go on to say that any random variable that, like a mean, is the sum of many small influences will have approximately a normal distribution. (This would explain why so many biological traits — height, weight, etc. — are normally-distributed.)