Hello, please help me with the following questions. thanks :)

#1

Use technology or a z-score table to answer the question.

The weights of boxes of rice produced at a factory are normally distributed with a mean of 24 ounces and a standard deviation of 1.3 ounces. Consider a shipment of 1200 boxes of rice.

How many of the boxes will weigh 25 ounces or less?

192

265

935

1008

#2:

Suppose the scores on a grammar quiz are normally distributed with a mean of 76 and a standard deviation of 8.

Which group describes 16% of the population of grammar quiz scores?

scores below 60

scores below 68

scores above 80

scores above 94

#3

The ages of people entering a pie eating contest are normally distributed with a mean of 35 years and a standard deviation of 8.2 years.

What is the age of a pie eating contestant with a z-score of −2.1 ?

#4

Use technology or a z-score table to answer the question.

The weights of books in a bookcase are normally distributed with a mean of 11.6 pounds and a standard deviation of 1.3 pounds. There are 136 books on the bookshelf.

How many of the books will weigh 10 pounds or more?

109

116

121

130

Guest Apr 19, 2019

#1**+1 **

#1

Use technology or a z-score table to answer the question.

The weights of boxes of rice produced at a factory are normally distributed with a mean of 24 ounces and a standard deviation of 1.3 ounces. Consider a shipment of 1200 boxes of rice.

How many of the boxes will weigh 25 ounces or less?

[ 25 - 24 ] / 1.3 = .77 = z score translates to 0.7794

So...about 78% will weight 25 ounces or less ≈ 936 [ 935 is the closest answer ]

CPhill Apr 19, 2019

#2**+1 **

#2:

Suppose the scores on a grammar quiz are normally distributed with a mean of 76 and a standard deviation of 8.

Which group describes 16% of the population of grammar quiz scores?

This will be all the scores less than 1 standard deviation* below* the mean.......so....this will represent all the scores < 76 - 8 = 68

CPhill Apr 19, 2019

#3**+1 **

#3

The ages of people entering a pie eating contest are normally distributed with a mean of 35 years and a standard deviation of 8.2 years.

What is the age of a pie eating contestant with a z-score of −2.1 ?

[ Age - 35 ] / [ 8.2] = -2.1 multiply both sides by 8.2

Age - 35 = -17.22 add 35 to both sides

Age ≈ 17.8 years

CPhill Apr 19, 2019

#5**+1 **

Use technology or a z-score table to answer the question.

The weights of books in a bookcase are normally distributed with a mean of 11.6 pounds and a standard deviation of 1.3 pounds. There are 136 books on the bookshelf.

How many of the books will weigh 10 pounds or more?

[ 10 - 11.6] / 1.3 = -1.23 z score which translates to .1093

So [ 1 - .1093] (136) ≈ 121 weigh 10 pounds or more

CPhill Apr 19, 2019