1. Find the probability that a household spent less than $25.00.
z = (“raw” score –mean) / standard deviation mu = 45.16, sigma = 10, z = (x - mu)/sigma
(a) z = (25 - 45.16)/10 = -2.016 or P (z < -2.016)
P(x < $25) = P(z < -2.016) = 0.0219
2. Find the probability that a household spent more than $50.00.
1 – NORMDIST (50, 45, 16, 10, TRUE)
(b) z = (50 - 45.16)/10 = 4.84/10 = 0.484
P(x > $50) + P(z > 0.484) = 0.3142
3. What proportion of the households spent between $30.00 and $40.00?
(c) z = (30 - 45.16)/10 = -1.516 and z = (40 - 45.16)/10 = -0.516
P($30 < x < $40) = P(-1.516 < z < -0.516) = 0.2382 OR 23.8% of the population spent between $30.00 and $40.00
4. 99% of households spent less than what amount?
z- score for 99% area is 2.3263
x = mu + z * sigma = 45.16 + 2.3263 * 10 = 99% of households spent less than $68.42
NORMINV (.99, 45.16)
References
Levine, D., Stephan, D., Krehbiel, T., & Berenson, M. (2008) Statistics for managers using Microsoft Excel w/cd. (5th ed.). Upper Saddle River,
References: Levine, D., Stephan, D., Krehbiel, T., & Berenson, M. (2008) Statistics for managers using Microsoft Excel w/cd. (5th ed.). Upper Saddle River, NJ