Respuesta :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 125, \sigma = 24[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 125}{24}[/tex]
[tex]X - 125 = 1.28*24[/tex]
[tex]X = 155.72[/tex]
The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.
Answer: 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
