Using the Z-table, we can determine the probability that a sample of 120 steady smokers spend between $19 and $21 is 0.986.
We have given that,
A sample is consider as survay to determine mean unt of steady smokers.
sample size (n) = 120
sample mean (X-bar ) = $20
standard deviations of sample (s) = $5
confidence interval is (19,21)
firstly we find the Z-Score value for given sample , C.I = X-bar +- Z(s/√n)
=> either 19 = 20 + Z(5/√120) or
21 = 20-Z(5/√220)
=> either Z = -0.456 or Z = 1/0.456
=> -2.19<Z<2.19
Using the either two tailed test or Z-table we can easily calculate the value of Probability.
the value of P( - 2.19<Z<2.19) is 0.986 ~ 0.99
Hence, the required probability is 0.986
To learn more about Probability or p -value , refer :
https://brainly.com/question/13786078
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