For Independent Events, P(A) × P(B) = P(A∩B)

so we have, P(A∩B) = 0.4×0.1 = 0.04

P(A') = 1 - 0.4 = 0.6

This information can be represented on a Venn diagram as shown below

P(A'∪B) means the union of everything that is not A with everything that is B

P(A'∪B) = 0.06 + 0.54 + 0.04 = 0.64

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2 years ago

In a certain year, when she was a high school senior, Idonna scored 671 on the mathematics part of the SAT. The distribution of

goldfiish [28.3K]

Answer:

Idonna's standardized score is 1.41.

Jonathan's standardized score is 0.55.

A.) Idonna's score is higher than Jonathan's

Step-by-step explanation:

Z-score:

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.

Idonna scored 671 on the mathematics part of the SAT. The distribution of SAT math scores in that year was Normal with mean 509 and standard deviation 115.

This means that her standardized score is Z when $X = 671, \mu = 509, \sigma = 115$ . So

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

$Z = \frac{671 - 509}{115}$

$Z = 1.41$

Idonna's standardized score is 1.41.

Jonathan took the ACT and scored 24 on the mathematics portion. ACT math scores for the same year were Normally distributed with mean 21.1 and standard deviation 5.3 .

This means that his standardized score is Z when $X = 24, \mu = 21.1, \sigma = 5.3$