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Helen [10]
3 years ago
14

The comprehensive strength of concrete is normally distributed with μ = 2500 psi and σ = 50 psi. Find the probability that a ran

dom sample of n = 5 specimens will have a sample mean diameter that falls in the interval from 2499 psi to 2510 psi.
Mathematics
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

The probability that the diameter falls in the interval from 2499 psi to 2510 psi is 0.00798.

Step-by-step explanation:

Let's define the random variable, X = "Comprehensive strength of concrete". We have information that X is normally distributed with a mean of 2500 psi and a standard deviation of  50 psi (or a variance of 2500 psi). In other words, X \sim N(2500, 2500).

We want to know the probability of the mean of X or \bar{X} that falls in the interval [2499;2510]. From inference theory we know that :

\bar{X} \sim N(2500, \frac{2500}{5}) \Rightarrow \bar{X} \sim N(2500,500)

Now we can find the probability as follows:

P(2499 \leq \bar{X} \leq 2510) \Rightarrow P(\frac{2499 - 2500}{500} \leq \frac{\bar{X} - 2500}{500} \leq \frac{2499 - 2500}{500} ) \Rightarrow\\\Rightarrow P(-0.002 \leq \frac{\bar{X} - 2500}{500} \leq 0.02 ) \Rightarrow P(-0.002 \leq Z \leq 0.02 )

Where Z \sim N(0,1), then:

P(-0.002 \leq Z \leq 0.02 ) \approx P(0 \leq Z \leq 0.02 ) = P(Z \leq 0.02 ) - P(Z \leq 0) \\P(0 \leq Z \leq 0.02 ) = 0.50798 - 0.5 = 0.00798

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In a group of 10 ​kittens, 5 are female. Two kittens are chosen at random. ​a) Create a probability model for the number of male
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Answer:

(b) Expected number of male kitten E(x) = 1

(c) Standard deviation is ±\frac{2}{3}.

Step-by-step explanation:

Given that,

Number of kittens in a group are 10. Out of these 5 are female.

To find:- (a) create a probability model of male kitten chosen.

    So,  total number of kitten = 10

           number of female kitten = 5    

then, Number of male kitten = 10-5= 5

Now total number of ways to choosing two kittens from group of 10 = ^{10} C_{2}

                                                                                                                 = \frac{10!}{2!\times8!}

                                                                                                                 = 45

for choosing (i) No male P(x=0) = P(Two female) =  \frac{^{5} C_{2}}{45}

                                                                           = \frac{2}{9}

                     (ii) 1 male P(x=1) = P(1 male and 1 female) = \frac{^{5} C_{1}\times^{5} C_{1}}{45} =  \frac{25}{45} =\frac{5}{9}

                     (iii) 2 male P(x=2)= P(2 male) =\frac{^{5} C_{2}}{45} =  \frac{2}{9}

      x_{i}                             0                     1                      2

P(X=x_{i})                           \frac{2}{9}                      \frac{5}{9}                      \frac{2}{9}

(b) what is expected number of male kittens chosen ?

    Expected number of male kitten E(x) =  o\times \frac{2}{9} + 1\times \frac{5}{9} + 2\times\frac{2}{9}

                                                                  = 1

(c) what is standard deviation ?

                              \sigma(x) = \sqrt{ (E(x^{2} )-(Ex)^{2})

No,                          

                               Ex^{2} =o\times \frac{2}{9} + 1\times \frac{5}{9} + 4\times\frac{2}{9}

                                       = \frac{13}{9}

                               \sigma(x)=\sqrt{\frac{13}{9} - 1 }=\sqrt{\frac{4}{9} }  

                                       = ± \frac{2}{3}

                               

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