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murzikaleks [220]
3 years ago
10

What is the probability of getting heads when flipping a coin and getting a number greater than or equal to 4 when rolling a sin

gle die?
Mathematics
2 answers:
lbvjy [14]3 years ago
5 0
"Assuming a fair coin<span> and a fair 6-sided </span>die<span>. </span>Coin<span> has 2 sides with </span>equal probability<span>, 50% each. </span>Die<span> has 6 sides with </span>equal probability<span>, 1/6 odds </span>for<span> each side. ... The</span>probability of getting heads<span> is 1/2, and the </span>probability of getting<span> 5 or 6 is 1/3, and so you simply multiply 1/2 x 1/3, which is 1/6."</span>
Sedaia [141]3 years ago
5 0

Answer's 1/4 in APEX

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She buys 2/8 more pound of granola than banana chips

Step-by-step explanation:

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Read 2 more answers
A population of plastic chairs in a factory has a weight's mean of 1.5 kg and a standard deviation of 0.1 kg . Suppose a sample
Firlakuza [10]

Answer:

0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

In this question, we have that:

\mu = 1.5, \sigma = 0.1, n = 100, s = \frac{0.1}{\sqrt{100}} = 0.01

What is the probability that the sample mean will be within +0.02 of the population mean?

Sample mean between 1.5 - 0.02 = 1.48 kg and 1.5 + 0.02 = 1.52 kg, which is the pvalue of Z when X = 1.52 subtracted by the pvalue of Z when X = 1.48. So

X = 1.52

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

By the Central Limit Theorem

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

Z = \frac{1.52 - 1.5}{0.01}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 1.48 ​

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

Z = \frac{1.48 - 1.5}{0.01}

Z = -2

Z = -2 has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

3 0
3 years ago
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