1. A company has been studying the performance of their sales department. The me
ID: 3207040 • Letter: 1
Question
1. A company has been studying the performance of their sales department. The measure of performance used is: average monthly sales amount in dollars. For the past 12 months, the average monthly sales amount has been $4,532 with a standard deviation of $386. Calculate the percentage of months that the average monthly sales amount exceeds $5,800.
2.
In a written test given to a large class comprising 42 students, the test scores were found to be normally distributed with a mean of 78 and a standard deviation of 7. A minimum score of 60 was needed to pass the test. A score of 90 or greater was needed to earn an A in the test. Answer the following questions;
a) What is the probability that a randomly selected student would pass the test?
b) What percentage of students earned an A?
3. A fastener company is supplying washers to a manufacturer of kitchen appliances who has specified that the outside diameter be 0.500 ± 0.025 inch. Measurements made on 250 washers show that the process is in good statistical control, the diameters are normally distributed, and the mean and standard deviation for the process are 0.505 and 0.0065 inch, respectively. What percent of the washers made by this process are within the specifications? Sketch the appropriate graph to represent the distribution of the process along with the specifications.
Explanation / Answer
from normal distribution z=(X-mean)/std deviation
1) P(X>5800)=1-P(Z<(5800-4532)/386)=1-P(Z<3.285)=1-0.99949=0.00051
2)a)for stduent to pass then test P(X>60)=1-P(X<60)=1-P(Z<(60-78)/7)=1-P(Z<-2.5714)=1-0.0051=0.9949
b) for student to get A grade P(X>90)=1-P(X<90)=1-P(Z<(90-78)/7)=1-P(Z<1.7143)=1-0.9568=0.0432
hence total number of student who got A grade =0.0432*42=1.8144
3)here P(0.475<X<0.525)=P((0.475-0.505)/0.0065<Z<(0.525-0.505)/0.0065)=P(-4.6154<Z<3.0769)=0.998954-0.0000=0.998952
hence 99.8952% will be Ok
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