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Tju [1.3M]
1 year ago
6

The likelihood that a patient with a heart attack dies of the attack is 0.04 (i.e., 4 of 100 die of the attack). Suppose we have

50 patients who suffer a heart attack. What is the probability that all will survive
Mathematics
1 answer:
Sonbull [250]1 year ago
6 0

The probability solved by binomial distribution that all will survive is 0.8154.

<h3>What is Binomial distribution? </h3>

When each trial has the same probability of achieving a given value, the number of trials or observations is summarized using the binomial distribution. The likelihood of observing a specific number of successful outcomes in a specific number of trials is determined by the binomial distribution.

Computation of probability of all survived people from heart attack;

A heart attack patient has a 0.04 percent chance of dying from the attack (i.e., 4 of 100 die of the attack).

What is the likelihood that each of the five patients who experience a heart attack will survive?

We'll refer to a victory in this scenario as a heart attack (p = 0.04). In other words, we are interested in the likelihood that none of our n=5 patients will die (0 successes).

Each attack has a chance of being fatal or not, with a probability of 4 percent for all patients, and each patient's result is independent.

Assume for the purposes of this example that the five individuals being examined are unrelated, of the same age, and free of any concomitant disorders.

By binomial distribution,

\begin{gathered}P(0 \text { successes })=\frac{5 !}{0 !(5-0) !} 0.04^{0}(1-0.04)^{5-0} \\P(0 \text { successes })=\frac{5 !}{5 !}(1)(0.96)^{5}=(1)(1)(0.8154)=0.8154\end{gathered}

Therefore, with a 4 % chance that anyone will die, there is an 81.54% chance that every patient will survive the onslaught. The outcomes in this example could be 0, 1, 2, 3, 4 or 5 successes (fatalities).

To know more about Binomial Distribution, here

brainly.com/question/27794898

#SPJ4

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. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person
bulgar [2K]

Answer:

2.28% probability that a person selected at random will have an IQ of 110 or greater

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 100, \sigma = 5

What is the probability that a person selected at random will have an IQ of 110 or greater?

This is 1 subtracted by the pvalue of Z when X = 110. So

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

Z = \frac{110 - 100}{5}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that a person selected at random will have an IQ of 110 or greater

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