mean = 35000
St.dev =5000
x = 40000
z = (x - mean)/sd/sqrt(n)
The incomes in a certain lg. population of college teachers have a normal distribution with a mean of $35,000 and st. dev. of $5000. Four teachers are randomly selected. What is the probability their average salary will exceed $40,000?
3 answers
n =
z = (40000-35000)/5000/sqrt(4)
z = 5000/(5000/2)
z = 2
P(z >2)
z = (40000-35000)/5000/sqrt(4)
z = 5000/(5000/2)
z = 2
P(z >2)
mean = 35000
st.dev = 5000
x = 40000
n = 4
z = (40000-35000)/5000/sqrt(4)
z = 5000/(5000/2)
z = 2
P(z >2) = 0.0228
st.dev = 5000
x = 40000
n = 4
z = (40000-35000)/5000/sqrt(4)
z = 5000/(5000/2)
z = 2
P(z >2) = 0.0228