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Artyom0805 [142]
3 years ago
6

Determine the range of values containing 90% of the population of x. From a very large data set (N > 50,000), x is found to h

ave a mean value of 192.0 units with a standard deviation of 10 units. Assume x is normally distributed.
Mathematics
1 answer:
Grace [21]3 years ago
3 0

Answer:

The range of values containing 90% of the population of x is from 175.55 units to 208.45 units.

Step-by-step explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

\u = 192, \sigma = 10

Determine the range of values containing 90% of the population of x.

The lower end of this interval is the value of X when Z has a pvalue of 0.5 - 0.9/2 = 0.05

The upper end of this interval is the value of X when Z has a pvalue of 0.5 + 0.7/2 = 0.95

Lower end

X when Z has a pvalue of 0.05, so X when Z = -1.645

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

-1.645 = \frac{X - 192}{10}

X - 192 = -1.645*10

X = 175.55

Upper end

X when Z has a pvalue of 0.95, so X when Z = 1.645

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

1.645 = \frac{X - 192}{10}

X - 192 = 1.645*10

X = 208.45

The range of values containing 90% of the population of x is from 175.55 units to 208.45 units.

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