# Business Analytics - Essay Example

The question asks for the probability of people who think that “Made in America” ads boost sales and utilizes social media. Please find my work below: P(0. 44 + 0. 78) = P(0. 44) + (0. 78) – (0. 85) this gives an answer as seen below: P (122) = P(122 – 85) =0. 37; therefore, the answer is there is a probability of 37% chance that people think that “Made in America” ads boost sales and utilize no social media. On the other hand, the second portion of this question utilizes the conditional probability steps: Probe (A given B) = Probe (A and B) / Probe(B).

To summarize event A is 0. 44 of people think that “Made in America” boosts sales, and (from answer 1 . A. ) 37% of people think that “Made In America” ads boost sales and utilize no social media. The question asks for the probability of people who think that “Made In America” ads boost sales given that no one utilizes social media. Please find my work below: P(O. 44 – 0. 37) = 0. 37. The next step I will subtract 1 from the 78% of people who classical media online. 1 -0. 78 = 022. Then I delve 0. 07 / 0. 22 0. 3282: therefore, round to the second decimal, the answer is there is a ND utilize no social media.

Question 2 This is problem had me slammed, if I am correct, I think this problem is a normal distribution problem. I will utilize the normal distribution steps: . Toastmaster the average monthly household cellular phone bill is $73 (mum); and has a standard deviation of $1 1. 35. Therefore, I will need to utilize a “Z” table to find the “z-score” for $100. The first portion of this problem asks for the probability that a randomly selected monthly cell phone bill is more than $100. To find this answer I utilize the “z” table. Therefore, x equals $100, and the rest of my work is below: mum)/s or 11. 5 = 2. 37885 or 2. 4, “Z” equals 0. 0087. This means that there is nearly a 9% chance that a monthly cell phone bill is more than $100. The second portion of this problem asks for the probability that a randomly selected monthly cell phone bill is between $60 and $83. Please find my work below: Z = (x- NIL’) / s 73)/ 11. 35 = -1. 14537 and mum)/s 73)/ 11. 35 = 0. 881057. To find the probability I would go back to the “Z” table and look up (-1. 14537 < Z < 0. 881057). The area to the right has 0. 874 and 0. 1891. roarnve at the final answer I would subtract: 0. 874 - 0. 1891 = 0. 849. So, there is a 68% chance that a monthly cell phone bill will be between $60 and $83. The next portion of this problem also looks for the probability utilizing the "Z" table. It asks for the probability that a randomly selected monthly cell phone bill is between $80 and $90. Therefore, my work shows below: Z = (x- mu) / s or Z = (80 - 73) / 11. 35 = 0. 61674 and Z = (x- mu) / s or Z 73) / 11. 35 = 1. 4978. Again, I go back to the "Z" table and look up 0. 61674 1. 4978. The area to the nght has 0. 2687 and 0. 671. -roarnve at the final answer I would subtract: 0. 2687 - 0. 671 = 0. 2016. So, there is a 20% chance that a monthly cell phone bill will be between $80 and $90. The final question asks for the probability that a randomly selected monthly cell phone bill is no more than $55. Utilizing the same methods in a, b, and c my answers follow below: Z = (x- mu) / s or Z = (55 - 73) / 11. 35 = -1. 5859; now utilizing the "Z" table I need to find the probability of Z < -1. 5859 = 0. 0564. Therefore, there is nearly a 6% chance that a monthly cell phone bill will be more than $55. Question 3 This final question is the same as the previous question.

This requires the “Z” answers. To summarize 950 randomly selected business travelers responded to a survey regarding their most recent business trips; 19% responded that it was for an internal company visit. So, my first question asks for the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit. My answers are below: First I determined my “P” (19%), “N” (950), and “Z” (25%). Then I got the Z score for 25%: Z = (P-P) / P) / N) or (0. 25 – 0. 19)/ (1 – 0. 19)/950) = 1. 907. The other equation P(p > 0. 25) = P(Z> 1. 4907) = 0. 0680, which means that there is a 6% or 7% chance that more than 25% of business travelers will say their most recent business trip will be an internal company visit. The second portion of this questions asks for the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit. To find this answer I 0. 19)/ (1 – 0. 19)/ 950)z = -3. 1427 and – (sort(p * (1 – P) / N). This revises the following equation: Z = (0. 0 – 0. 19) / sort(O. 19 * (1 – 0. 19) / 950) = z = 0. 7857. Again, I utilize the “Z” tables to find the probability of -3. 1427 < Z < 0. 7857. This gives an answer of 0. 7832. This means that there is a 78% percent probability chance that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit. The final question asks for the probability that between 133 and 171 of the business travelers say that the reason for their most recent business trip was an internal company visit.

Utilizing the same steps, I will find the “P” value for 133 and 171: P(133) = 133 / 950 = 0. 14 and P(171)= 171 / 950 = 0. 18 Now I will utilize the “Z” tables and find the answer for these: / (sort(p * (1 – P) / N). Z-(0. 14- 0. 19)/ (1 – 0. 19)/950) = -3. 9284 and Z-(0. 18- 0. 19)/ (1 – 0. 19)/950) = -0. 7857. NOW find the probability of -3. 9284 and 07857 from the “Z” tables. They show -3. 9284 < Z < -0. 7857 = 0. 216. Therefore, there is a 21% chance that between 133 and 171 business travelers say that the reason for their most recent business trip was an internal company visit.