Probability question. Find the probability that the data set falls within 45% to 52% of the data setone of your employees has cautioned that your enterprise expand a brand new product. A survey is designed to have a look at whether or not or not there may be hobby inside the new product. The reaction is on a 1 to five scale with 1 indicating without a doubt could no longer buy, · · ·, and 5 indicating absolutely could purchase. For an preliminary analysis, you will document the responses 1, 2, and 3 as No, and 4 and 5 as yes.a. five people are surveyed. what is the opportunity that as a minimum three of them replied sure?b. 100 people are surveyed. what's the approximate possibility that between 45% to fifty two% of people answered sure?For component a) There are 5 choices that human beings can respond by way of 1, 2 ,three ,4 and five. when you consider that 1, 2 and three are considered "No", the possibility of a person answering "No" is three/five. For choices four and 4, the probability of a person responding with this is 2/5.This seems like it fallows a binomial distribution so I calculated the chance of P(three) + P(four) + P(five).but for component b), i'm stressed. i will calculate the chance of a person pronouncing sure but I do not know how to calculate the chance that a percentage of people announcing sure. Does absolutely everyone recognize how to technique this question? I though approximately the use of the Z table, however that already calculates region.

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Probability question. Find the probability that the data set falls within 45% to 52% of the data setone of your employees has cautioned that your enterprise expand a brand new product. A survey is designed to have a look at whether or not or not there may be hobby inside the new product. The reaction is on a 1 to five scale with 1 indicating without a doubt could no longer buy, · · ·, and 5 indicating absolutely could purchase. For an preliminary analysis, you will document the responses 1, 2, and 3 as No, and 4 and 5 as yes.a. five people are surveyed. what is the opportunity that as a minimum three of them replied sure?b. 100 people are surveyed. what&#039;s the approximate possibility that between 45% to fifty two% of people answered sure?For component a) There are 5 choices that human beings can respond by way of 1, 2 ,three ,4 and five. when you consider that 1, 2 and three are considered &quot;No&quot;, the possibility of a person answering &quot;No&quot; is three/five. For choices four and 4, the probability of a person responding with this is 2/5.This seems like it fallows a binomial distribution so I calculated the chance of P(three) + P(four) + P(five).but for component b), i&#039;m stressed. i will calculate the chance of a person pronouncing sure but I do not know how to calculate the chance that a percentage of people announcing sure. Does absolutely everyone recognize how to technique this question? I though approximately the use of the Z table, however that already calculates region.

Antwan Wiley · Accepted Answer

Let X denote the number of people answering yes. Then X is a bionomial random variable.Let n denote the number of trials (i.e. people surveyed). Let p denote the probability of a yes-response on a given trial, and let q denote the probability of a no-response. Then, as you&#039;ve argued, we have  p  =  2      /    5  ,    q  =  3      /    5.Then in part (a), the probability of at least three people answering yes is  P  (  x  =  3  )  +  P  (  x  =  4  )  +  P  (  x  =  5  )  =  …  .So far so good!Now for part (b):You see, 45 of 1000 is 450, and 52 of 1000 is 520. So I reckon the probability you want is                    P        (        X        =        450        )        +        P        (        X        =        451        )        +        ⋯        +        P        (        X        =        520        )                            =                  ∑                      r            =            450                                520                          P        (        X        =        r        )                                          =                  ∑                      r            =            450                                520                                                                              (                                                    1000              r                                                      )                                                                          (                          2              5                        )                    r                                      (                          3              5                        )                                1000            −            r                                                            =        ⋯

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