We have a normal distribution with mean = 4.2 and sd =1.3.
a. So the probability x>5 is given by:
1-norm.cdf(5, mean = 4.2, sd = 1.3) = 1-0.73085 = 0.26915
or equivalently, p(z>(5-4.2)/1.3).
b. We have a sample mean = 5, sample size n = 8. Since 8 is very small we use the t-distribution to estimate sample mean distributions, with df = n-1 = 7.
The standard error of our sample mean is sd/sqrt(n)
So our t-stat is given by (sample mean - population mean) / standard error
t = (5-4.2) / 0.49
Therefore, the probability our sample mean is greater than 5 will be one minus the probability it is less, which is
1 - P(t = 1.628, df = 7)
= 1 - 0.926241741
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