25. A major department store has determined that its customers charge an average of $500 per month with a standard deviation of $80. Assume the amounts of charges are normally distributed.
a. What percentage of customers charges more than $380 per month?
P(x > 380) = 1- p(x<380)
(P(x<380) = z value = x - mean / std dev
Z Value = (380 - 500)/80 = -1.5
P(x<380) = 0.0668 How do I get this number? I get how to get the z value but where does the .0668 come from? I looked on my table in the book but I can't find .0668 anywhere.
P(x > 380) = 1- p(x<380)
= .9331 or 93.31%
Solution Summary
The solution investigates the normal distribution data from a department store.
... a sample of 100 is taken from a population which has a normal distribution with a mean of 80 and a standard deviation of 40. What is the probability that the ...
1. This week we learned about normal distributions: a. Using ... a normal distribution and a standard normal distribution. ... values to z scores to find probability? ...
... Critical value = one tail T distribution 99% confidence ... The expert examines probability, normal distributions, and binomial probabilities for randomly ...
... 1 year or less, that is determine the probability that x ... 3))=P(-0.54<Z<0.54)=0.4108 from normal table ... means, and draw a dotplot for sampling distribution of the ...
... means that the height of the probability function can in ... the sum of all the probabilities must equal ... How would you apply the normal distribution to facilitate ...
... exponential probability distribution D. Poisson probability distribution. ... 9. Revising probabilities when new information ... form, the normal distribution: A. has a ...
... spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a ...
... If the probability distribution is not known, we cannot calculate probability. Uses Excel for calculating probabilities in a Normal Distribution. ...
... Please see the attached file. The solution gives the detailed answers to probability, normal distribution, z score and confidence interval problems. ...
