How many area codes (ABC) would be possible if the first 2 digits can be any number 2-9 & last digit can be any number 0-9?

1 answer:

8 0

The answer would be A. 640

Possibilities for the first two are 8 and 8 (2-9 is 8 numbers) and the last is 10 (0-9 is 10 numbers)

So the probability is found by multiplying 8*8*10 which is 640

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e. 0.977

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:

$\mu = 478, \sigma = 64$