# Applied Statistics for Business Decision Making

Pages: 2 (281 words) Published: March 20, 2011
In a recent year, about two-thirds of U.S. households purchased ground coffee, with a mean of \$45.16 and a standard deviation of \$10.00. The probability of the household spending more than \$50.00, less than \$25.00, and spending between \$30.00 and \$40.00, is highly likely. The principles of probability reveal the long-tern predictability and experiment short-term results or outcomes which occur in situations (Levine, Stephan, Krehbiel, & Berenson, 2008).

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, NJ

References: Levine, D., Stephan, D., Krehbiel, T., & Berenson, M. (2008) Statistics for managers using Microsoft Excel w/cd. (5th ed.). Upper Saddle River, NJ