4.1_HW_Excel_Notes

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Apr 3, 2024

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Quetsion 1 Fair Die 1 Odd 3 Odd 2 Even 6 sides so… = 3/6 Sides 3 Odd 4 Even 3 Even 5 Odd 6 Even Question 2 1 2 3 * Combination w/ all Girls (GGG) is include 1 BBB BBG BGG 2 GBG BGB 4 Combinations w/ at most 1 Boy 3 GBB GGB GGG so… Question 3 1 2 3 4 5 6 1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 5 Combina Find the probability an odd number is thrown in the toss of a fair die. (Enter as a decimal) Suppose a family has three children . Find the probability they have at most one boy . (Enter as a decimal) 8 Different Combinations Consider the tossing of a pair of dice. Find the probability that the sum of the two die is 6 . (Round to the hundre
2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 34 total Co 6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Question 4 1 2 3 4 5 6 1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 21 Combin 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 34 total Co 6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Question 5 Grade Frequency A 5 5 Students received an A B 9 C 9 so… =5/35 D 7 35 Students Total in the class F 5 Question 6 Grade Frequency A 5 23 Students Passed the Class B 9 C 9 so… = 23/35 D 7 35 Students Total in the class F 5 Question 7 Grade Frequency A 5 12 Students Failed the Class B 9 so... = 12/35 Consider again the tossing of a pair of dice. Find the probability the sum is greater than 6 . (Round to the hundre Referring to the table calculate the probability a randomly selected student received an A in the course. (Round Refer to the table. Given that receiving at least a C is passing the course, calculate the probability a student passe Refer to the table. Calculate the probability a student did not pass the course . (Round to the hundredths)
C 9 D 7 35 Students Total in the class F 5 Questions 8 Selling Range Frequency 20 homes between given frequency $250,000 - $299,999 10 $300,000 - $349,999 20 so... $350,000 - $399,999 20 $400,000 - $450,000 10 60 Total homes Question 9 * Includes home ranges $250,00 Selling Range Frequency $250,000 - $299,999 10 30 homes between given frequency $300,000 - $349,999 20 $350,000 - $399,999 20 so... $400,000 - $450,000 10 60 Total homes Question 10 Calculate the probability a randomly selected home sold in the range of $350,000 to $399,999 . (Round to the hu Suppose you only want to spend at most $349,999 on a home. Calculate the probability you will find a home in y A single card is drawn from a standard deck of 52 playing cards . Find the probability the card is a numbered card
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Question 11 Question 12 A single card is drawn from a standard deck of 52 playing cards . Find the probability the card is not a spade . A single card is drawn from a standard deck of 52 playing cards . Find the probability the card is not a king, quee
Question 13 Question 14 Remember An exam distribution has a normal distribution with mean 75 and standard deviation 10 . Calculate the probabili In this prob The 1st dev MCAT scores are normally distributed with an average score of 500.5 and a standard deviation of 10.5. Calculat
In this prob The proble You have t
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Answer = 0.5 ed too b/c: 0 Boys ≤ 1 Boy Answer = 4/8 = 0.5 ations edths)
Anwser = 5/34 = 0.14 ombinations nations Anwser so… = 21/34 = 0.58 ombinations Anwser = 0.14 Anwser = 0.66 Anwser = 0.34 edths) to the hundredths) ed the course.
Anwser = 20/60 = 0.33 00 to $349,999 Anwser = 30/60 = 0.5 Non Numbered Cards 12 Cards that are Kings, Queens, and Jacks 4 Aces in the deck 52 - (12 + 4) = 36 so... Anwser = 36/52 = 0.69 The Total Amount of Cards is 52 undredths) your price range. (Round to the hundredths) d. (Round to the hundredths)
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13 Cards that are Spades 52 - 13 = 39 Anwser so… = 39/52 = 0.75 The Total Amount of Cards is 52 12 Cards that are Kings, Queens, and Jacks 52 - 12 = 40 Anwser so… = 40/52 = 0.77 en, or jack . (Round to the hundredths)
The Total Amount of Cards is 52 2. We determine the the range for each Deviation: Anwser 68 r that 1st, 2nd, and 3rd Deviations are a percentage and each Deviation is just a segment of the data found in th ity that if the exam is given to a group of students that a student scores between 65 and 85 . (Write as a whole n blem, we draw and fill out the bell curve first w/ an average of 75 1. Given Standard Deviation = 10 = so from statring from 500.5 , we count by every point by 10 1st = 68% = 65 to 85 2nd = 95% = 55 to 95 3rd = 99.7 % = 45 to 105 viation ( 68% ) contains the scores bewteen 65 to 85 te the probability you get a score higher than 521.5 if you take the exam. (Don't include the percent sign.) Ex:) 1st Deviation = 68% = 68 of 100 , so the 1st Deviation is only 68% of the data found in the graph
2. We determine the the range for each Deviation: *this is shaded in blue in the graph 5/2 Anwser = 2.5 blem, we draw and fill out the bell curve first w/ the given average of 500.5 1. Given Standard Deviation = 10.5 = so from statring from 500.5 , we count by every point by 10.5 1st = 68% = 490 to 511 2nd = 95% = 479.5 to 521.5 3rd = 99.7 % = 469 to 532 em asks scores higher than 521.5 , so we are looking for everything past 521.5 only 1. 521.5 is the highest number within the 2nd Deviation 2. 2nd Deviation = 95% = 95 of 100 or 95% of the data found in the graph 3. 100% - 95% = 5% , which is the range outside of the 2nd Deviation 4. The problem is only asking for everything past 521.5 , which is only the other half of the graph = half of 5% 5 . Half of 5% = the probability of 2.5% of getting a score higher than 521.5 for the MCAT Exam
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he bell curve graph number . For example 95% would just be input as 95 ) h