Question

1. A nutritionist is looking at the connection between hours of TV watched and choice of sugary snacks in children identified as at risk for obesity. He asks children to document the number of hours of TV they watch and the number of sugary snacks they eat each day as shown in the first three columns of the following table.

Child	Hours of TV watched (X)	Number of sugary snacks eaten (Y)
A	3	3
B	1	3
C	2	1
D	4	2
E	5	4

a. What type of research design is represented in this study (experimental? Non-experimental? Correlational?). Explain your answer

b. What do you think the independent variable is? What is the scale of measurement?

c. What do you think the dependent variable is? What is the scale of measurement?

d. For hours of TV watched, calculate the mean and standard deviation and Z score for each child

e. For number of sugary snacks eaten, calculate the mean and standard deviation and Z score for each child

f. Comparing the Z scores for each child (on TV and snacks), do you notice any trend in the data? Describe.

2. For a normal distribution with a mean µ=80 and σ=20, find the proportion of the population corresponding to each of the following scores:

• Scores greater than 85
• Scores less than 100
• Scores between 70 and 90

3. Rochester, New York, averages µ=21.9 inches of snow for the month of December. The distribution of snowfall amounts is normal with a standard deviation of σ=6.5 inches. This year the local jewelry store is advertising a refund of 50% off all purchases made in December if we finish the month with more than 36 (3ft) of snow. What is the probability that the jewelry store will have to pay off on its promise?

4. Why do we convert raw scores in Z scores?

Answer 1

2)

a)

µ =    80                  
σ =    20                  
                      
P ( X ≥   85   ) = P( (X-µ)/σ ≥ (85-80) / 20)              
= P(Z ≥   0.25   ) = P( Z <   -0.250   ) =    0.4013   (answer)

b)

P( X ≤    100   ) = P( (X-µ)/σ ≤ (100-80) /20)      
=P(Z ≤   1.00   ) =   0.84134   (answer)

c)

we need to calculate probability for ,                                      
P (   70   < X <   90   )                      
=P( (70-80)/20 < (X-µ)/σ < (90-80)/20 )                                      
                                      
P (    -0.500   < Z <    0.500   )                       
= P ( Z <    0.500   ) - P ( Z <   -0.50   ) =    0.6915   -    0.3085   =    0.3829   (answer)

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3)

µ =    21.9                  
σ =    6.5                  
                      
P ( X ≥   36   ) = P( (X-µ)/σ ≥ (36-21.9) / 6.5)              
= P(Z ≥   2.17   ) = P( Z <   -2.169   ) =    0.0150   (answer)

4)

because z score allows us to calculate the probability of a score in normal distribution

and enables us to compare two scores that are from different normal distributions.