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True [87]
3 years ago
13

Test scores are normally distributed with a mean of 500. Convert the given score to a z-score, using the given standard deviatio

n. Then find the percentage of students
who score below 620, if the standard deviation is 80.​
Mathematics
2 answers:
Anna71 [15]3 years ago
6 0

Answer:

The z score formula is given by:

z = \frac{620-500}{80}= 1.5

And if we use the normal standard distribution table or excel we got:

P(z

And the percentage would be 93.3%

Step-by-step explanation:

Let X the random variable that represent the test scores of a population, and for this case we know the distribution for X is given by:

X \sim N(500,80)  

Where \mu=500 and \sigma=80

We are interested on this probability

P(X

The z score formula is given by:

z=\frac{x-\mu}{\sigma}

The z score formula is given by:

z = \frac{620-500}{80}= 1.5

And if we use the normal standard distribution table or excel we got:

P(z

And the percentage would be 93.3%

Bond [772]3 years ago
3 0

Answer:

The percentage of students who scored below 620 is 93.32%.

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 question, we have that:

\mu = 500, \sigma = 80

Percentage of students who scored below 620:

This is the pvalue of Z when X = 620. So

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

Z = \frac{620 - 500}{80}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

The percentage of students who scored below 620 is 93.32%.

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