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2 years ago

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In a test of a printed circuit board using a random test pattern, an array of 16 bits is equally likely to be 0 or 1. Assume the

bits are independent.
(a) What is the probability that all bits are 1s? Round your answer to six decimal places (e.g. 98.765432).
(b) What is the probability that all bits are 0s? Round your answer to six decimal places (e.g. 98.765432).
(c) What is the probability that exactly 8 bits are 1s and 8 bits are 0s? Round your answer to three decimal places (e.g. 98.765).

1 answer:

Doss [256]2 years ago

8 0

Answer:

A) b) and c) answer has been explained below.

Step-by-step explanation:

For the question would use 1's and 0's and taking probabilities to be equal for both using random test pattern whose formula is when p = q

then Simplify to

P[k] = nCk /2^n

A.Probability that all bits are 1s

16c16/2^16 = 1/65536

B. Probability that all bits are 0s

16c0/2^16 = 1/65536

C.the probability that exactly 8 bits are 1s and 8 bits are 0s

16c8/2^16 = 12870/65536 =>0.1963 ≈ 19.63%

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Step-by-step explanation:

Data given and notation

$\bar X=5.0611$ represent the sample mean

$\sigma=0.2803$ represent the population standard deviation for the sample

$n=42$ sample size

$\mu_o =5$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:

Null hypothesis: $\mu = 5$

Alternative hypothesis: $\mu \neq 5$

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(z>1.413)=0.158$

Conclusion

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