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garik1379 [7]
4 years ago
7

A bag contains 20 white marbles and 30 black marbles. if 7 marbles are chosen, what is the probability that there will be 2 whit

e marbles and 5 black marbles?
Mathematics
1 answer:
AlekseyPX4 years ago
8 0
This is a hypergeometric distribution problem.
Population (N=50=W+B) is divided into two classes, W (W=20)  and B (B=30).
We calculate the probability of choosing w (w=2) white and b (b=5) black marbles.
Hypergeometric probability gives
P(W,B,w,b)=C(W,w)C(B,b)/(C(W+B,w+b)
where
C(n,r)=n!/(r!(n-r)!) the number of combinations of choosing r out of n objects.

Here
P(20,30,2,5)
=C(20,2)C(30,5)/(20+30, 2+5)
=190*142506/99884400
=0.2710

Alternatively, doing the combinatorics way:
#of ways to choose 2 from 20 =C(20,2)
#of ways to choose 5 from 30=C(30,5)
total #of ways = C(50,7)
P(20,30,2,5)=C(20,2)*C(30,5)/C(50,7)
=0.2710 
as before.
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We have to round the number 83.5851 to the nearest hundredth.
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8 0
3 years ago
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An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y
aksik [14]

Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                               P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average time = 23.92 minutes

             s = sample standard deviation = 6.72 minutes

             n = sample of times = 40

             \mu = population mean assembly time

<em> Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, a 98% confidence interval for the population mean, </u>\mu<u> is; </u>

P(-2.426 < t_3_9 < 2.426) = 0.98  {As the critical value of z at 1%  level

                                               of significance are -2.426 & 2.426}  

P(-2.426 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.426) = 0.98

P( -2.426 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

P( \bar X-2.426 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

<u>98% confidence interval for</u> \mu = [ \bar X-2.426 \times {\frac{s}{\sqrt{n} } } , \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ]

                                     = [ 23.92-2.426 \times {\frac{6.72}{\sqrt{40} } } , 23.92+2.426 \times {\frac{6.72}{\sqrt{40} } } ]  

                                    = [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

7 0
3 years ago
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Answer: True

Step-by-step explanation:

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san4es73 [151]
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