Question #192281

a normally distributed population has mean 57.7 and standard deviation 12.1

a) find the probability that a single randomly selected element x of the population is less than 45

b) find the mean and the standard deviation of x-x for sample size 16

c) find the probability that the mean of sample of size 16 drawn from the population is less than 45


1
Expert's answer
2021-05-13T01:16:26-0400

mean(μ)=57.7mean\left(\mu \right)=57.7

Standarddeviation(σ)=12.1000Standard\:deviation\left(\sigma \right)=12.1000


(a). Probability that a single randomly selected element X of the population is less than 45 :

P(X<45)=(Z<(4557.7)12.1)=P(Z<1.05)=0.1469\:P\left(X<45\right)=\left(Z<\frac{\left(45-57.7\right)}{12.1}\right)=P\left(Z<-1.05\right)=0.1469


(b).n=16n=16

mean of sampling distribution μx=57.7\mu x=57.7

standard deviation of sampling distribution σx=σn=3.025\sigma \overline{x}=\frac{\sigma }{\sqrt{n}}=3.025


(c). Probability that the mean of a sample of size 16 drawn from this population is less than 45

P(X<45)=(Z<(4557.7)3.025)=P(Z<4.2)=0.0000P\left(X<45\right)=\left(Z<\frac{\left(45-57.7\right)}{3.025}\right)=P\left(Z<-4.2\right)=0.0000



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