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12345 [234]
3 years ago
10

You randomly observe purchases at a candy store and record the weight of each bag of candy with 100 different customers. You fin

d the proportion of weights greater than 2 pounds. Is investigating this sample proportion the same as investigating the sampling distribution of the proportion?
Mathematics
2 answers:
laila [671]3 years ago
3 0
<span>No, in this case the two components in question would be different. On one hand, investigating the sample proportion refers to the specific amount that is seen to be distributed on average. However, with sampling distribution, this refers more specifically to how much is distributed on a long-term basis.</span>
katrin2010 [14]3 years ago
3 0

Solution:

→→When you are talking about sample proportion You are considering a Single ,set of same kind of Data,Like body weight of  Students in a class, from which you are taking out samples for Observation.

For example, A colony is being studied whether each human in that colony is healthy or not by checking their body mass index.

→→When we are investigating or considering Sampling Distribution we consider two different sets or kinds of Data set Possessing Similar Characteristics because we have to compare these two sets of Data.

For example , Take two colonies A and B , Check their honesty by observing how many floors they have made in their Plot.Whether they care or don't care about their Surrounding or Mother Earth by looking, if number of floors exceeds than 2 i.e ground floor and first floor , it means Dishonest , if maximum number of floor =2 ,it means honest.

No, investigating this sample proportion is not  same as investigating the sampling distribution of the proportion.


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Answer: 0.5

Step-by-step explanation:

Binomial probability formula :-

P(x)=^nC_x\ p^x(q)^{n-x}, where P(x) is the probability of getting success in x trials , n is the total trials and p is the probability of getting success in each trial.

Given : The probability that the adults follow more than one game = 0.30

Then , q= 1-p = 1-0.30=0.70

The number of adults surveyed : n= 15

Let X be represents the adults who follow more than one sport.

Then , the probability that fewer than 4 of them will say that football is their favorite sport,

P(X\leq4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)\\\\=^{15}C_{0}\ (0.30)^0(0.70)^{15}+^{15}C_{1}\ (0.30)^1(0.70)^{14}+^{15}C_{2}\ (0.30)^2(0.70)^{13}+^{15}C_{3}\ (0.30)^3(0.70)^{12}+^{15}C_{4}\ (0.30)^4(0.70)^{11}\\\\=(0.30)^0(0.70)^{15}+15(0.30)^1(0.70)^{14}+105(0.30)^2(0.70)^{13}+455(0.30)^3(0.70)^{12}+1365(0.30)^4(0.70)^{11}\\\\=0.515491059227\approx0.5

Hence, the probability rounded to the nearest tenth that fewer than 4 of them will say that football is their favorite sport =0.5

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