Question

1. The reading speed of Ms. Alayeb's second grade students is approximately normal, with a mean 60 WPM and a standard deviation of 10 WPM. (Please round your answer to 4 decimal places )

a) What is the probability a randomly selected student will read more than 65 WPM?

b)What is the probability that a random sample of 27 Ms. Alayeb's second grade students results in a mean reading rate of less than 56 WPM?

c)What is the probability that a random sample of 18 Ms. Alayeb's second grade students results in a mean reading rate of more than 63 WPM?

2. Slices of pizza for a certain brand of pizza have a mass that is approximately normally distributed with a mean of 66.5 grams and a standard deviation of 2.34 grams.

a) For samples of size 18 pizza slices, what is the standard deviation for the sampling distribution of the sample mean?

b) What is the probability of finding a random slice of pizza with a mass of less than 66.1 grams?

c) What is the probability of finding a 18 random slices of pizza with a mean mass of less than 66.1 grams?

d) What sample mean (for a sample of size 18) would represent the bottom 15% (the 15th percentile)?

Answer 1

Here we have : = 60, = 10

Let x be the reading speed of Ms. Alayeb's second grade students is approximately normal.

a) Here we need to find,  probability that a randomly selected student will read more than 65 WPM.

p ( x > 65 )

= p ( z > 0.5 )

= 1 - p ( z 0.5 )

= 1 - 0.6915 ------------ ( using excel formula " =norm.s.dist(0.5,1)" )

= 0.3085

b)

Here n = 27, we need to find, probability that students results in a mean reading rate of less than 56 WPM.

p ( < 56 )

= p ( z < -2.08 )

= 0.0188   ------------ ( using excel formula " =norm.s.dist(-2.08,1)" )

c) Here n= 18, we need to find, probability that students results in a mean reading rate of more than 63 WPM.

p ( > 63 )

= p ( z > 1.27)

= 1- p ( z 1.27 )     

= 1 - 0.8980   ------------ ( using excel formula " =norm.s.dist(1.27,1)" )

= 0.1020