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umka21 [38]
3 years ago
5

The pie chart to the right shows how adults rate their financial shape. Suppose 4 people are chosen at random from a group of 11

00. what is the probability that all four would rate their financial shape as excellent? (make the assumption that the 1100 people are represented by the pie chart)
PIE CHART DATA:
Excellent 7%
Good 44%
Fair 35%
Poor 13%
Other 1%
Mathematics
2 answers:
vekshin13 years ago
8 0

Answer:

Step-by-step explanation:

Given that a pie chart shows data as

Excellent 7%

Good 44%

Fair 35%

Poor 13%

Other 1%

If 4 people are chosen at random we find that each person is independent of other and there are two outcomes either excellent or not excellent.

p = Prob for success = p = 7%=0.07

q = Prob for non success = 1-0.07 = 0.93

If x is a random variable representing the excellent persons then X is binomial with n =4, p = 0.07

Required probability = P(X=4)

=p^4\\=(0.07)^4\\=0.00002401

Over [174]3 years ago
3 0

Answer:

0.000022304

Step-by-step explanation:

Total number of people = 1100

Number of people chosen = 4

Excellent percentage is 7% or 0.07

So, number of people rated excellent = 0.07\times1100= 77

Others who did not choose excellent = 1100-77=1023

So, p(randomly choose 4 excellent)= \frac{(77C4)(1023C0)}{1100C4}

= \frac{77C4}{1100C4}

=1353275/60671970975 = 2.2304e^{-5}

= 0.000022304

Or second and easy method is simply multiplying:

\frac{77}{1100} \times\frac{76}{1099} \times\frac{75}{1098} \times\frac{74}{1097}

= 32478600/1456127303400 = 0.000022304

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Suppose that 75% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider grou
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Answer:

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c) 0.2251

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Step-by-step explanation:

We are given the following information:

We treat trucks undergoing a brake inspection passin as a success.

P( trucks undergoing a brake inspection passes the test) = 75% = 0.75

a) Conditions for binomial probability distribution

  1. There are n independent trial.
  2. Each trial have a success probability p
  3. The probability of success is same for all trials.

Then the number of trucks undergoing a brake inspection follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 15

b) P(proportion of groups will between 8 and 10 trucks pass the inspection)

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P(8\leq x \leq 10) = P(x = 8) + P(x = 9) + P(x = 10)\\= \binom{15}{8}(0.75)^8(1-0.75)^7 + \binom{15}{9}(0.75)^9(1-0.75)^6 + \binom{15}{10}(0.75)^{10}(1-0.75)^5\\= 0.0393 + 0.0917 + 0.1651\\= 0.2961

c) P( exactly 3 trucks fail the inspection)

p = 0.25

P(x = 3)\\= \binom{15}{3}(0.25)^3(1-0.25)^{12}\\=0.2251

d) Mean and standard deviation

\mu = np = 15(0.75) = 11.25\\\sigma = \sqrt{np(1-p)} = \sqrt{15(0.75)(1-0.75)} = 1.667

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