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3 years ago

15

The mean time it takes for workers at a factory to insert a delicate bolt into an engine is 15 minutes. The standard deviation o

f time to insert the bolt is 4.0 minutes and the distribution of time is approximately Normally distributed. For a randomly selected worker, what is the approximate probability the bolt will be inserted in 19 minutes or less

1 answer:

faust18 [17]3 years ago

6 0

Answer:

required probability is 0.8413

Step-by-step explanation: Given mean $\mu =15$ minutes, standard deviation $\sigma =4$ seconds, we need to find $P\left ( X\leq 19 \right )$ . Using z score

$z=\frac{X-\mu }{\sigma }\\=\frac{19-15}{4}\\=1$

So,

$P\left ( X\leq 19 \right )\\=P\left ( Z\leq 1 \right )\\=0.8413$

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Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

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Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

i.

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

$P(Atleast\ 2 \ girls) = 0.5$

iii.

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Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

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