Question

Suppose that we take a random sample of size n from a population having a mean μ and a standard deviation of σ. What is the probability or confidence we have that our sample will be within +/- 1 of the population mean μ?

(a) μ = 10, σ = 2, n = 25

Answer 1

Solution:

We are given

µ = 10

σ = 2

n = 25

We have to find P(10 – 1 < Xbar < 10 + 1) = P(9<Xbar<11)

P(9<Xbar<11) = P(Xbar<11) – P(Xbar<9)

Find P(Xbar<11)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (11 – 10)/[2/sqrt(25)]

Z = 1/0.4

Z = 2.5

P(Z< 2.5) = P(Xbar<11) = 0.99379

(by using z-table)

Now find P(Xbar<9)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (9 – 10)/[2/sqrt(25)]

Z = -1/0.4

Z = -2.5

P(Z< -2.5) = P(Xbar<9) = 0.00621

(by using z-table)

P(9<Xbar<11) = P(Xbar<11) – P(Xbar<9)

P(9<Xbar<11) = 0.99379 - 0.00621

P(9<Xbar<11) = 0.98758

Required probability = 0.98758