A firm has invoices which follows a normal distribution with a mean of $2.090 and a standard deviation of $150.
a. What is the probability that the invoice will be between $2000 and $2200?
b. What percentage of invoices will be over $2260?
c. Above what amount will 90% of the invoices lie?
Solution:
Given:
Mean = 2090
SD = 150
Part a
P(2000<X<2200) = P(X<2200) – P(X<2000)
Z = (X – mean) / SD
Z = (2200 - 2090)/150
Z = 0.733333
P(Z<0.733333) = 0.768322
(by using z-table)
Z = (2000 - 2090)/150
Z = -0.6
P(Z<-0.6) = 0.274253
(by using z-table)
P(2000<X<2200) = P(X<2200) – P(X<2000)
P(2000<X<2200) = 0.768322 - 0.274253
P(2000<X<2200) = 0.494069
Required probability = 0.494069
Part b
Here, we have to find P(X>2260) = 1 – P(X<2260)
Z = (X – mean) / SD
Z = (2260 - 2090)/150
Z =1.133333
P(Z<1.133333) = 0.871463
(by using z-table)
P(X>2260) = 1 – P(X<2260)
P(X>2260) = 1 – 0.871463
P(X>2260) = 0.128537
Required probability = 0.128537
Required percentage = 12.85%
Part c
X = mean + Z*SD
Z for 90% area = 1.281552
(by using z-table)
X = 2090 + 1.281552*150
X = 2282.233
Required amount = $2,282.23
A firm has invoices which follows a normal distribution with a mean of $2.090 and a...
A firm has invoices which follows a normal distribution with a mean of $2.090 and a standard deviation of $150. a. What is the probability that the invoice will be between $2000 and $2200? b. What percentage of invoices will be over $2260? c. Above what amount will 90% of the invoices lie? please show details with this part of the question.
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