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Greeley [361]
3 years ago
9

If there is an 80% chance of rain and a 50% chance of wind and rain, what is the probability that it is windy, given that it is

rainy? Round your answer to the nearest percent.
Mathematics
2 answers:
ankoles [38]3 years ago
6 0
If the total probability when given it is rainy is 80% and the windy probability among 80% is 50% the total probability is...

50% / 80% = 62.5% ≈ 63%


Pie3 years ago
5 0

Answer:

The conditional probability that it is windy, given that it is rainy is is 63%.

Step-by-step explanation:

It is given that there is 80% chance of rain, that is P(A)=0.80

50%chance of wind and rain, that is P(A∩B)=0.50

Now, the conditional probability that it is windy, given that it is rainy is:

P(B|A)=\frac{P(A{\cap}B)}{P(A)}

P(B|A)=\frac{0.50}{0.80}

P(B|A)=0.625

Thus, the conditional probability that it is windy, given that it is rainy is is 63%.

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Express 112 as a product of prime numbers in index form?
alexgriva [62]
                                   112
                                    /  \
                                  2   56
                                       /  \
                                     2   28
                                          /  \
                                         2  14
                                              /  \
                                             2  7

So, the answer is 2 x 2 x 2 x 7 or 2^3 x 7
6 0
3 years ago
The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c
Anestetic [448]

Answer:

a) The mean is \mu = 60

b) The standard deviation is \sigma = 9

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be 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.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when X = 55.5, Z = -0.5

So

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

-0.5 = \frac{55.5 - \mu}{\sigma}

-0.5\sigma = 55.5 - \mu

\mu = 55.5 + 0.5\sigma

The probability a student selected at random takes no more than 71.52 minutes to complete the examination equals 0.8997.

This means that when X = 71.52, Z has a pvalue of 0.8997. This means that when X = 71.52, Z = 1.28

So

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

1.28 = \frac{71.52 - \mu}{\sigma}

1.28\sigma = 71.52 - \mu

\mu = 71.52 - 1.28\sigma

Since we also have that \mu = 55.5 + 0.5\sigma

55.5 + 0.5\sigma = 71.52 - 1.28\sigma

1.78\sigma = 71.52 - 55.5

\sigma = \frac{(71.52 - 55.5)}{1.78}

\sigma = 9

\mu = 55.5 + 0.5\sigma = 55.5 + 0.5*9 = 55.5 + 4.5 = 60

Question

The mean is \mu = 60

The standard deviation is \sigma = 9

6 0
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Answer:

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Step-by-step explanation:

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