Stats-random samples?

Rayovac type D batteries "have a continuous use life" that is approximately normally distributed with mean 21 hrs. and standard dev 1.38 hrs. Duracell type D batteries have a "continuous use life" that is approximately normally distributed with mean 24 hrs and standard dev 2.13 hrs.





FIND THE PROBABILITY (to four places after the decimal) THAT:


[a]...a randomly selected rayovac lasts longer than a randomly selected duracell.





[b]...given simple random samples of four each, the rayovak batteries have a longer mean "continuous use life"





[c]...a single rayovak lasts under 22 hrs





[d]...the mean "continuous use life" of 4 randomly selected rayovak batteries is less than 22 hrs





[e]...the mean "continuous use life" of 16 randomly selected rayovak batteries is less than 22 hrs

Stats-random samples?
x=Rayovac :mu1=21 sd1=1.38


y=Duracell: mu2=24 sd2=2.13


a) p(x %26gt; y) = p(x-y %26gt; 0)


u=x-y has a normal distribution with mean -3, variance=(1.38)^2+(2.13)^2=1.9044+4.5369... or sd=2.5378


p( u %26gt; 0) = p(z %26gt; (0+3)/2.5378)=p(z %26gt; 1.1821)=0.119


b)n1=4 n2=4


H0: mu1=mu2


H1: mu1 %26gt; mu2


compute sd as


sqrt[s1^2/n1+s2^2/n2] -- (1)


same as (a) but use sd from (1)


We do not have the sample means here.


c)p( x %26gt; 22)=p( z %26gt; (22-21)/1.38)=p(z %26gt;0.7246)=0.2358


d)sd=1.38/sqrt(4)=0.69


p(x %26lt; 22)= p( z %26lt; (22-21)/0.69)=p(z %26lt; 1.449)=0.9265


e)sd=1.38/sqrt(16)=1.38/4=0.345


p( x %26lt; 22)= p( z %26lt; (22-21)/0.345)=p(z %26lt; 2.1)=0.9821

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