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3 years ago

14

Find the range of f(x)=x+3 when the domain (2,4,6)

1 answer:

jarptica [38.1K]3 years ago

5 0

<u>f(x) = x + 3</u>
f(2) = 2 + 3
f(2) = 5

f(4) = 4 + 5
f(4) = 9

f(6) = 6 + 5
f(6) = 11

{(2, 5), (4, 9), (6, 11)}

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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.

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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

3 years ago