Mathematics, 20.07.2021 01:00 brenton14

Consider the probability that at most 85 out of 136 DVDs will work correctly. Assume the probability that a given DVD will work correctly is 52%. Specify whether the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

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Answer from: erinolson07cats

Since both $np \geq 10$ and $n(1-p) \geq 10$ , the necessary conditions are satisfied.

0.9945 = 99.45% probability that at most 85 out of 136 DVDs will work correctly.

Step-by-step explanation:

Test if the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

It is needed that:

$np \geq 10$ and $n(1-p) \geq 10$

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume the probability that a given DVD will work correctly is 52%.

This means that p = 0.52

136 DVDs

This means that n = 136

Test the conditions:

$np = 136*0.52 = 70.72 \geq 10$

$n(1-p) = 136*0.48 = 65.28 \geq 10$

Since both $np \geq 10$ and $n(1-p) \geq 10$ , the necessary conditions are satisfied.

Mean and standard deviation:

$\mu = E(X) = np = 136*0.52 = 70.72$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{136*0.52*0.48} = 5.83$

Consider the probability that at most 85 out of 136 DVDs will work correctly.

Using continuity correction, this is $P(X \leq 85 + 0.5) = P(X \leq 85.5)$ , which is the p-value of Z when X = 85.5. So

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

$Z = \frac{85.5 - 70.72}{5.83}$

Z = 2.54

Z = 2.54 has a p-value of 0.9945.

0.9945 = 99.45% probability that at most 85 out of 136 DVDs will work correctly.

Answer from: Quest

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0

the value of x is equal to 0.

Answer from: Quest

7.3

step-by-step explanation:

81+30=111180 minus 111=691/2×19=9.569÷9.5=7.263157895~7.3
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