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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answered: 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.

answered: Guest

Letter b sorry i’m i’m wrong xx

answered: Guest

The answer is 10. and if you go by pemdas you would 1. multiplication 2. division 3. addition 4. subtraction
Solve the following problem using the order of operations. then match each operation with its place

Solve the following problem using the order of operations. then match each operation with its place

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