The manufacturer of holiday lights claims that 96% of his strings of lights are
ID: 3365748 • Letter: T
Question
The manufacturer of holiday lights claims that 96% of his strings of lights are free of defects. Assume a store purchases 200 strings of lights, what is the probability that more than 195 strings will be free of defects?
A local motel has 100 rooms. The occupancy rate for the winter months is 60%. Find the probability that in a given winter month fewer than 70 rooms will be rented. Use the normal distribution to approximate the binomial distribution.
In a Fall semester, students purchase a mean of $375 worth of textbooks with a standard deviation of $45. If 125 students are surveyed, what proportion of the students will purchase less than $385 worth of books?
A survey was conducted to determine the number of hours people listen to the radio. While most of the listening is in the car, it turns out that the mean is 2.5 hours with a standard deviation of 0.75 hours. If 49 people are surveyed, what is the probability they listen to less than 2.25 hours of radio?
In a Fall semester, students purchase a mean of $525 worth of textbooks with a standard deviation of $75. If 164 students are surveyed, what proportion of the students will purchase between $515 and $530 worth of books?
Explanation / Answer
Question 1
Pr(string of lights are free of defects) = 0.96
Number of string of lights = 200
Expected Number of lights who dont have defects = 200 * 0.96 = 192
Standard deviation of lights who dont have defects = sqrt [ 200 * 0.96 *0.04] = 2.7713
Pr(X > 195) = BIN (X > 195.5 ; 192; 2.7713) [ using continuity correction]
Z = (195 .5 - 192)/ 2.7713 = 1.263
Pr(X > 195) = NORM(X > 195.5 ; 192; 2.7713) = 1 - Pr(Z < 1.263) = 1 - 0.8967 = 0.1033
Question 2
Occupancy rate in winter = 0.60
Total number of Motel rooms = 100
Expected number of Occupied room = 100 * 0.60 = 60
Standard deviation of Occupied rooms = sqrt [100 * 0.60 * 0.40] = 4.899
Pr(X < 70) = NORM (X < 69.5 ; 60 ; 4.899) [ by continuity correction]
Z = (69.5 - 60)/ 4.899 = 1.9392
Pr(X < 70) = Pr(Z < 1.9392) = 0.9738
Question 3
Students purchases of notebooks worth = $ 375
Standard deviation of purchases of notebooks worth = $ 45
Standard error of the sample mean se0= /sqrt(n) = 45/ sqrt(125) = 4.025
so,
Pr(X < $ 385) = NORM (X < $ 385 ; $ 375 ; $4.025)
Z = (385 - 375)/4.025 = 2.4845
Pr(X < $ 385) = NORM (X < $ 385 ; $ 375 ; $4.025) = Pr(Z < 2.4845) = 0.9935
so 99.35% of students will purchase less than $ 385 worth of books.
Question 4
Mean listening hours music for radio = 2.5 hours
Standard deviation hours music for radio = 0.75 hours
Standard error of the sample mean = /sqrt(n) = 0.75/ sqrt(49) = 0.75/7 = 0.1071
Pr(x < 2.25) = Pr(x < 2.25; 2.5 ; 0.1071)
Z = (2.25 - 2.5)/ 0.1071 = -2.33
Pr(x < 2.25) = Pr(Z < -2.33) = 0.0098
Question 5
Students purchases of notebooks worth = $ 525
Standard deviation of purchases of notebooks worth = $ 75
sample size = 164
standard error of the sample mean se0= /sqrt(n) = 75/ sqrt(164) = 5.8565
so,
Pr($ 515 < X < $ 530) = NORM (X < $ 530 ; $ 525 ; $5.8565) - NORM (X < $ 515 ; $ 525 ; $5.8565)
= (Z2) - (Z1)
where is the standard normal cumulative distribution function
Z2= (530 - 525)/ 5.8565 = 0.8538
Z1= (515 - 525)/ 5.8565 = -1.7075
Pr($ 515 < X < $ 530) = (Z2) - (Z1) = (0.8538) - (-1.7075) = 0.8034 - 0.0439 = 0.7595
SO 75.95% of the students will purchase between $515 and $530 worth of books
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.