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Luda [366]
1 year ago
6

Thirty-two percent of workers in Troberca City are college graduates. You randomly select 50 workers and ask them is he or she i

s a
college graduate. Find the probability that less than 18 workers are college graduate. (Approximating a Binomial Distribution)
(A) 32.6%
B 67.4%
64.0%
36.0%
Mathematics
2 answers:
pychu [463]1 year ago
8 0
B around 68% would be college graduates
Gnoma [55]1 year ago
3 0

The correct option is B) 67.4%

<h3>What is the z- score table?</h3>

A standard normal table (also called the unit normal table or z-score table) is a mathematical table for the values of ϕ, indicating the values of the cumulative distribution function of the normal distribution. Z-Score, also known as the standard score, indicates how many standard deviations an entity is, from the mean.

Given here, P(probability that a worker is a college graduate) = 32%

                                                                                                      = 0.32

q (probability that a worker is not a college graduate) is 1-0.32 = 0.68  

now μ(mean) = n×p

                      = 50×0.32

                      = 16

Again standard deviation σ = \sqrt{npq\\}

                                           σ = \sqrt{50\cdot 0.32\cdot 0.68}

                                           σ = 3.2985

now P(x<17.5) = P((x-μ)/σ <(17.5-16)/3.2985)

                      = P(z<0.45)  

                      = 0.674 [using z table]

                      = 67.4%

Hence option 2 is the correct answer

Learn more about z-table here:

brainly.com/question/29283457

#SPJ2

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So, Tina was wrong, she had to divide by 3x.

Answer:

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A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
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Ivahew [28]

Answer:

\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.

Step-by-step explanation:

<u>Perfect squares</u>: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ...

To find \sf \sqrt{75} , identify the perfect squares immediately <u>before</u> and <u>after</u> 75:

  • 64 and 81

\begin{aligned}\sf As\;\; 64 < 75 < 81\; & \implies \sf \sqrt{64} < \sqrt{75} < \sqrt{81}\\&\implies \sf \;\;\;\;\;8 < \sqrt{75} < 9 \end{aligned}

\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.

See the attachment for the correct placement of \sf \sqrt{75} on the number line.

6 0
11 months ago
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