The scores of students on a standardized test are normally distributed with a mean of 300 and a standard deviation of 40. (a) what proportion of scores lie between 220 and 380 points?

Sagot :

0.9544 = 95.44% of scores lie between 220 and 380 points.

Normal distribution problems can be solved using the Z-score formula.

With a set of means and standard deviations, the Z-score for measure X is given by: After finding the Z-score, look at the Z-score table to find the p-value associated with that Z-score. This p-value is the probability that the value of the measure is less than X. H. Percentile of X. Subtract 1 from the p-value to get the probability that the value of the measure is greater than X.

[tex]z = \frac{x - \mu}{\sigma} \,[/tex]

We are given mean 300, standard deviation 40.

This means that µ= 300, σ = 40

What proportion of scores lie between 220 and 380 points?

This is the p-value of Z when X = 380 subtracted by the p-value of Z when X = 220.

X = 380

[tex]z = \frac{x - \mu}{\sigma} \,[/tex]

Z= (380-300)/40

Z= 2

Z=2 has a p-value of 0.9772.

X=300

[tex]z = \frac{x - \mu}{\sigma} \,[/tex]

Z= (220-380)/40

Z=-2

Z=-2 has a p-value of 0.9772.

0,9772 - 0,0228 = 0,9544

0.9544 = 95.44% of scores lie between 220 and 380 points.

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