I need answer for this problem

A complicated mechanical system contains 8 components. If there is a 18% chance that each component will fail during processing,

kicyunya [14]

Answer:

(a) The probability that fewer than 3 components will fail is 0.8392.

(b) The mean likely number of failures is 1.44.

Step-by-step explanation:

We are given that a complicated mechanical system contains 8 components. If there is an 18% chance that each component will fail during processing, and components fail independently of each other.

Let X = Number of components fail during processing

The above situation can be represented through the binomial distribution;

$P(X=r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ;x=0,1,2,.....$

where, n = number of samples (trials) taken = 8 components

r = number of success = fewer than 3 components will fail

p = probability of success which in our question is the probability

that each component will fail during processing, i.e. p = 18%

SO, X ~ Binom(n = 8, p = 0.18)

(a) The probability that fewer than 3 components will fail is given by = P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{8}{0}\times 0.18^{0} \times (1-0.18)^{8-0}+ \binom{8}{1}\times 0.18^{1} \times (1-0.18)^{8-1}+ \binom{8}{2}\times 0.18^{2} \times (1-0.18)^{8-2}$

= $1 \times 1 \times 0.82^{8}+ 8 \times 0.18^{1} \times 0.82^{7}+28 \times 0.18^{2} \times 0.82^{6}$

= 0.8392

(b) The mean likely number of failures is given by the following formula;

Mean of X, E(X) = n $\times$ p

= $8 \times 0.18$ = 1.44

Hence, the mean likely number of failures is 1.44.

3 0

3 years ago

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podryga [215]

I cant see ur attached thing

3 0

3 years ago

In an effort to cut costs and improve profits, any US companies have been turning to outsourcing. In fact, according to Purchasi

My name is Ann [436]

Answer:

a) 0.06% probability that 338 or more companies outsourced some part of their manufacturing process in the past two or three years.

b) 90.15% probability that 285 or more companies outsourced some part of their manufacturing process in the past two or three years.

c) 0.23% probability that 48% or less of these companies outsourced some part of their manufacturing process in the past two or three years.

Step-by-step explanation:

For questions a and b, the normal approximation to the binomial is used, while for question c, the central limit theorem is used.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

54% of companies surveyed outsourced some part of their manufacturing process in the past two to three years.

This means that $p = 0.54$

555 of these companies are contacted.

This means that $n = 555$

Mean and standard deviation: Normal approximation to the binomial:

$\mu = E(X) = np = 555*0.54 = 299.7$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{555*0.54*0.46} = 11.74$

a. What is the probability that 338 or more companies outsourced some part of their manufacturing process in the past two or three years?

Using continuity correction, this is $P(X \geq 338 - 0.5) = P(X \geq 337.5)$ , which is 1 subtracted by the p-value of Z when X = 337.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{337.5 - 299.7}{11.74}$

$Z = 3.22$

$Z = 3.22$ has a p-value of 0.9994.

1 - 0.9994 = 0.0006

0.0006*100% = 0.06%

0.06% probability that 338 or more companies outsourced some part of their manufacturing process in the past two or three years.

b. What is the probability that 285 or more companies outsourced some part of their manufacturing process in the past two or three years?

Using continuity correction, this is $P(X \geq 285 - 0.5) = P(X \geq 284.5)$ , which is 1 subtracted by the p-value of Z when X = 284.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{284.5 - 299.7}{11.74}$

$Z = -1.29$

$Z = -1.29$ has a p-value of 0.0985.

1 - 0.0985 = 0.9015

0.9015*100% = 90.15%

90.15% probability that 285 or more companies outsourced some part of their manufacturing process in the past two or three years.

c. What is the probability that 48% or less of these companies outsourced some part of their manufacturing process in the past two or three years?

Now we use the sampling distribution of the sample proportions, which have:

$\mu = p = 0.54$