48. Reading Rates The reading speed of sixth-grade students is approximately normal, with a mean speed of 125 words per minute and a standard deviation of 24 words per minute. (a) What is the reading speed of a sixth-grader whose reading speed is at the 90th percentile

The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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:

What is the reading speed of a sixth-grader whose reading speed is at the 90th percentile

This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.

The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.

the reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.84 per minute

Step-by-step explanation:

Since the the reading speed of sixth-grade students is approximately normal, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = reading speed

µ = mean speed

σ = standard deviation

From the information given,

µ = 125 words per minute

σ = 24 words per minute

Looking at the normal distribution table, the z value corresponding to the 90th percentile(0.9), is 1.285

Therefore,

1.285 = (x - 125)/24

24 × 1.285 = x - 125

30.84 = x - 125

x = 125 + 30.84

x = 155.84

1

step-by-step explanation:

it's the only multiple of both and the least