Unformatted text preview: ' 1 The distribution of weights of milk bottles 15 normally distributed with a mean of l. I lbs and a standard dviation (o)=0. 20 What as the probability that the WM will be
t th 0.99 lb 9
greaer an s. P(i7-.‘olcl) _.2 Y‘ 6 First Calculate the mean and standard error for the sampling distribution of a random sample of 5 milk bottles
0 By the CLT, is approximately normal with meah=1.1 and _ l-l
. standard error— — 0.2/JE =0.0894 " sham . O‘b‘M‘ >
. P( r >099)= 0.89 T - - -
. . . ~PC 7-: l' 7-33
m Closing prices of stocks have a right skewed distribution with a mean (u) of ﬁgs and 63.90. What IS the probability that the mean of a randomsamplegfﬁﬂ stocks will be less than $20? ‘ 11-?5‘f 3:231 Calculate the mean and standard error for the sampling distribution of a random sample of 40 stocks
0 By the CLT, a? is approximately normal with mean'=25 and
. standard error = 20/ m =3.1623 . P( r <20)=0.06 > Standard errors have exact alues depending on parameter values, e. g.,
I p(I — p) / n for a sample proportion - 0/5 for a sample mean > In practice, these parameter values are unknown. Inference methods use standard errors that
substitute sample values for the parameters in the exact formulas above These estimated standard errors are the numbers we use in practice > For a binomial random variable with r1 trials and probabilityp of success for each, the sampling distribution of the roportimmmas . o ean=p 0 Standard error = remnants”; n sunrﬂrmﬂlb'
> T ese values can e ound by taking the mean up and the standard deviation np(1 - p) for the binomial distribution of the number of successes and dividing by n An automobile' Insurer has found that repair claims have a mean of $920 and a standard deviation of
$870. Suppose that the next 100 claims can be regarded as a random sample from the long-run claims process.
What Is the probability that the average of the 100 claims 15 larger than $900? ...
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- Spring '08