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dangina [55]
3 years ago
10

During the assembly of a certain piece of exercise equipment, 3 welds are required during different phases of the assembly proce

ss. It is assumed that the welds are performed independently of one another. Due to variation in the materials used and in technique, only 97% of all welds are performed satisfactorily, with the remaining 3% being classified as defective. What percentage of all pieces of this type of equipment will have at least one defective weld
Mathematics
1 answer:
Karolina [17]3 years ago
5 0

Answer:

99.9973 %

Step-by-step explanation:

This is a binomial probability distribution.

Since the probability of satisfactory welds = 97% = 0.97 and the probability of defective welds = 3% = 0.03.

Since there are 3 welds and we require at least one being defective, our binomial probability is

P(x ≥ 1) = 1 - P(x ≤ 0) = 1 - P(0) = 1 - ³C₀(0.03)³(0.97)⁰ = 1 - 1 × ‭0.000027‬ × 1 = 1 - 0.000027 = 0.99997‬3 × 100% = 99.9973%

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A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

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For this exercise it is necessary to remember the following:

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Knowing this, and having the following expression given in the exercise:

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