A random sample of size n = 66 is taken from a population with mean = 12.5 and s

Question

A random sample of size n = 66 is taken from a population with mean = 12.5 and standard deviation = 6. Use Table 1.

a. Calculate the expected value and the standard error for the sampling distribution of the sample mean. (Negative values should be indicated by a minus sign. Round "expected value" to 1 decimal place and "standard error" to 4 decimal places.) Expected value and Standard error

b. What is the probability that the sample mean is less than 13? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.) Probability

c. What is the probability that the sample mean falls between 13 and 12? (Do not round intermediate calculations. Round "z" value to 2 decimal places and final answer to 4 decimal places.) Probability

Explanation / Answer

Mean is -12.5 and s is 6, thus standard error ,SE is s/sqrt(N)=6/sqrt(66)=0.7385 . z is given as (x-mean)/SE

a) Expected value is -12.5 and standard error is 0.7385

b) P(xbar<-13) =P(z<(-13-(-12.5))/0.7385)=P(z<-0.68) or 1-P(z<0.68), from normal distribution table we get 1-0.7517 =0.2483

c) P(-13<z<-12)=P((-13-(-12.5))/0.7385<z<(-12-(-12.5))/0.7385)=P(-0.68<z<0.68) or 2*P(z<0.68)-1 =2*0.7517-1=0.5034

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.