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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

 Apr 19, 2019
 #1
avatar+101838 
+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 ]

 

 

cool cool cool

 Apr 19, 2019
 #2
avatar+101838 
+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

 

 

cool cool cool 

 Apr 19, 2019
 #3
avatar+101838 
+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

 

 

cool cool cool

 Apr 19, 2019
 #4
avatar
0

I also forgot to add in this question:

 

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

The expression  P(z<1.33) represents the area under the standard normal curve below the given value of z.

What is  P(z<1.33)?

 

0.0918

0.0934

0.9066

0.9082

Guest Apr 19, 2019
 #5
avatar+101838 
+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

 

 

cool cool cool

 Apr 19, 2019

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