3) The reading speed of sixth-grade students is approximately normal, with a mea

Question

3) 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. [25 points total) (Chapter 7.2) a) What is the probability that a randomly selected sixth-grade student reads less than 100 words per minute? b) What is the probability that a randomly selected sixth-grade student reads more than 140 words per minute c) What is the probability that a randomly selected sixth-grade student reads between 110 and 130 words per minute? d) What is the reading speed of a sixth-grader whose reading speed is at the 90th percentile? e) A school psychologist wants to determine reading rates for unusual students (both slow and fast). Determine the reading rates of the middle 95% of all sixth grade students. What are the cutoff points for unusual readers?

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 125
standard Deviation ( sd )= 24
a.
LESS THAN
P(X < 100) = (100-125)/24
= -25/24= -1.0417
= P ( Z <-1.0417) From Standard Normal Table
= 0.1488
b.
GREATER THAN
P(X > 140) = (140-125)/24
= 15/24 = 0.625
= P ( Z >0.625) From Standard Normal Table
= 0.266
c.
BETWEEN THEM
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 110) = (110-125)/24
= -15/24 = -0.625
= P ( Z <-0.625) From Standard Normal Table
= 0.266
P(X < 130) = (130-125)/24
= 5/24 = 0.2083
= P ( Z <0.2083) From Standard Normal Table
= 0.5825
P(110 < X < 130) = 0.5825-0.266 = 0.3165
d.
P ( Z < x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.281552
P( x-u/s.d < x - 125/24 ) = 0.9
That is, ( x - 125/24 ) = 1.281552
--> x = 1.281552 * 24 + 125 = 155.757238
e.
LOWER/BELOW
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.959964
P( x-u/s.d < x - 125/24 ) = 0.025
That is, ( x - 125/24 ) = -1.959964
--> x = -1.959964 * 24 + 125 = 77.960864
UPPER/TOP
P ( Z > x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is 1.959964
P( x-u / (s.d) > x - 125/24) = 0.025
That is, ( x - 125/24) = 1.959964
--> x = 1.959964 * 24+125 = 172.039136

95% of the grade students will be [ 77.960864, 172.039136]

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