In: Statistics and Probability
In a recent year, Delaware had the highest per capita annual income with $51,803. Assume that σ = $4,850. A random sample of 39 state residents were selected.
What is the distribution of the sample mean income?
What is the probability that the sample mean income is greater
than $50,800?
Solution :
Given that ,
mean = = 51803
standard deviation = = 4850
n = 39
1)
= = 51803 and
= / n = 4850/ 39 = 776.6215
2)
P( > 50800) = 1 - P( < 50800)
= 1 - P(( - ) / < (50800 - 51803) / 776.6215)
= 1 - P(z < -1.29)
= 1 - 0.0985 Using standard normal table.
= 0.9015
Probability = 0.9015