What are the similarities and differences in solving literal equations and solving numerical equations?
2 answers:
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Given:
μ = 25 mpg, the population mean
σ = 2 mpg, the population standard deviation
If we select n samples for evaluation, we should calculate z-scores that are based on the standard error of the mean.
That is,
The random variable is x = 24 mpg.
Part (i): n = 1
σ/√n = 2
z = (24 -25)/2 = -0.5
From standard tables,
P(x < 24) = 0.3085
Part (ii): n = 4
σ/√n = 1
z = (24 -25)/1 = -1
P(x < 24) = 0.1587
Part (iii): n=16
σ/√n = 0.5
z = (24 - 25)/0.5 = -2
P(x < 24) = 0.0228
Explanation:
The larger the sample size, the smaller the standard deviation.
Therefore when n increases, we are getting a result which is closer to that of the true mean.
μ = 25 mpg, the population mean
σ = 2 mpg, the population standard deviation
If we select n samples for evaluation, we should calculate z-scores that are based on the standard error of the mean.
That is,
The random variable is x = 24 mpg.
Part (i): n = 1
σ/√n = 2
z = (24 -25)/2 = -0.5
From standard tables,
P(x < 24) = 0.3085
Part (ii): n = 4
σ/√n = 1
z = (24 -25)/1 = -1
P(x < 24) = 0.1587
Part (iii): n=16
σ/√n = 0.5
z = (24 - 25)/0.5 = -2
P(x < 24) = 0.0228
Explanation:
The larger the sample size, the smaller the standard deviation.
Therefore when n increases, we are getting a result which is closer to that of the true mean.