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An electrical company claims that the average life of the bulbs it manufactures is 1 300 hours with a standard deviation of 260 hours. If a random sample of 110 bulbs is chosen, what is the probability that the sample mean will be: a. greater than 1 160 hours? b. less than 1 260 hours? c. between 1 160 and 1 260 hours?

An electrical company claims that the average life of the bulbs it manufactures is 1 300 hours with a standard deviation of 260 hours. If a random sample of 110 bulbs is chosen, what is the probability that the sample mean will be: a. greater than 1 160 hours? b. less than 1 260 hours? c. between 1 160 and 1 260 hours?

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.2: Expected Value And Variance Of Continuous Random Variables
Problem 10E
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Example The average number of Facebook (FB) friends of the Senior High School learners in South Cotabato is 2, 896 with a standard deviation of 17. Assume that the variable is normally distributed. (a) If a learner is selected, what is the probability that he or she has more than 2 900 FB friends? (b) If 69 learners are randomly selected, what is the probability that their average FB friends is less than 2, 900? Solution to (a) Solution to (b) Solution 1. Identify the given u = 2 896, o = 17 X = 2 900, n=69 P (X <2 900) Steps Steps 1. Identify the given information. Solution µ = 2 896 a = 17 X = 2 900 P (X > 2 900) information. 2. Identify what is 2. Identify what is asked for. 3. Identify the formula to be asked for. 3. Identify the formula to be used. z = z-쁘 (to standardize 3000 since we're dealing with individual data) In 4. Solve the problem. used. 4. Solve the problem. 2 900 - 2896 z=- 17 2900 – 2896 17 = 0.24 (using the z-table, area to the left of the curve) P(X>2900)=P(X>0.24) -1 - 0.5948 V69 =1.95 (using the z-table, area to the left of the curve) P (X > 2 900) P (X > 1.95) =1- 0.97441= 0.03 So, the probability that the average number of FB friends of 69 =0.4052 5. State the final So, the probability 5. State the final answer. answer. that a student has more than 2 900 FB friends is 40.52%. randomly selected learners is 3%.
Transcribed Image Text:Example The average number of Facebook (FB) friends of the Senior High School learners in South Cotabato is 2, 896 with a standard deviation of 17. Assume that the variable is normally distributed. (a) If a learner is selected, what is the probability that he or she has more than 2 900 FB friends? (b) If 69 learners are randomly selected, what is the probability that their average FB friends is less than 2, 900? Solution to (a) Solution to (b) Solution 1. Identify the given u = 2 896, o = 17 X = 2 900, n=69 P (X <2 900) Steps Steps 1. Identify the given information. Solution µ = 2 896 a = 17 X = 2 900 P (X > 2 900) information. 2. Identify what is 2. Identify what is asked for. 3. Identify the formula to be asked for. 3. Identify the formula to be used. z = z-쁘 (to standardize 3000 since we're dealing with individual data) In 4. Solve the problem. used. 4. Solve the problem. 2 900 - 2896 z=- 17 2900 – 2896 17 = 0.24 (using the z-table, area to the left of the curve) P(X>2900)=P(X>0.24) -1 - 0.5948 V69 =1.95 (using the z-table, area to the left of the curve) P (X > 2 900) P (X > 1.95) =1- 0.97441= 0.03 So, the probability that the average number of FB friends of 69 =0.4052 5. State the final So, the probability 5. State the final answer. answer. that a student has more than 2 900 FB friends is 40.52%. randomly selected learners is 3%.
АCTIVITY An electrical company claims that the average life of the bulbs it manufactures is 1 300 hours with a standard deviation of 260 hours. If a random sample of 110 bulbs is chosen, what is the probability that the sample mean will be: a. greater than 1 160 hours? b. less than 1 260 hours? c. between 1 160 and 1 260 hours?
Transcribed Image Text:АCTIVITY An electrical company claims that the average life of the bulbs it manufactures is 1 300 hours with a standard deviation of 260 hours. If a random sample of 110 bulbs is chosen, what is the probability that the sample mean will be: a. greater than 1 160 hours? b. less than 1 260 hours? c. between 1 160 and 1 260 hours?
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ISBN:
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