The reading speed of second grade students in a large city is approximately norm

Question

The reading speed of second grade students in a large city is approximately normal, mean of 90 wpm and standard deviation of 10. A) what is probability a randomly selected student will read more than 94 wpm? B) what is probability that a random sample of 14 students result in a mean rate of more than 94 wpm? C) what is probability that a random sample of 28 students result in a mean rate of more than 94 wpm? D) how does increasing sample size effect probability? The reading speed of second grade students in a large city is approximately normal, mean of 90 wpm and standard deviation of 10. A) what is probability a randomly selected student will read more than 94 wpm? B) what is probability that a random sample of 14 students result in a mean rate of more than 94 wpm? C) what is probability that a random sample of 28 students result in a mean rate of more than 94 wpm? D) how does increasing sample size effect probability? A) what is probability a randomly selected student will read more than 94 wpm? B) what is probability that a random sample of 14 students result in a mean rate of more than 94 wpm? C) what is probability that a random sample of 28 students result in a mean rate of more than 94 wpm? D) how does increasing sample size effect probability?

Explanation / Answer

NORMAL DISTRIBUTION
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 ) = 90
standard Deviation ( sd )= 10

a.
P(X > 94) = (94-90)/10
= 4/10 = 0.4
= P ( Z >0.4) From Standard Normal Table
= 0.3446
NORMAL DISTRIBUTION
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/sqrt(n) ~ N(0,1)
mean ( u ) = 90
standard Deviation ( sd )= 10
sample size (n) = 14

b.
P(X > 94) = (94-90)/10/ Sqrt ( 14 )
= 4/2.673= 1.4967
= P ( Z >1.4967) From Standard Normal Table
= 0.0672

c.
P(X > 94) = (94-90)/10/ Sqrt ( 28 )
= 4/1.89= 2.1166
= P ( Z >2.1166) From Standard Normal Table
= 0.0171

d.
increasing sample size effect probability
the above two b,c probability values decreasing order because sample sizes increses

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