Answer:
81.85%
Step-by-step explanation:
Given :
The average summer temperature in Anchorage is 69°F.
The daily temperature is normally distributed with a standard deviation of 7°F .
To Find:What percentage of the time would the temperature be between 55°F and 76°F?
Solution:
Mean = 
Standard deviation = 
Formula : 
Now At x = 55
At x = 76
Now to find P(55<z<76)
P(2<z<-1)=P(z<2)-P(z>-1)
Using z table :
P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185
Now percentage of the time would the temperature be between 55°F and 76°F = 
Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.
One of the major advantage of the two-condition experiment has to do with interpreting the results of the study. Correct scientific methodology does not often allow an investigator to use previously acquired population data when conducting an experiment. For example, in the illustrative problem involving early speaking in children, we used a population mean value of 13.0 months. How do we really know the mean is 13.0 months? Suppose the figures were collected 3 to 5 years before performing the experiment. How do we know that infants haven’t changed over those years? And what about the conditions under which the population data were collected? Were they the same as in the experiment? Isn’t it possible that the people collecting the population data were not as motivated as the experimenter and, hence, were not as careful in collecting the data? Just how were the data collected? By being on hand at the moment that the child spoke the first word? Quite unlikely. The data probably were collected by asking parents when their children first spoke. How accurate, then, is the population mean?
Answer:1/2=0.5
3/4=0.75
7/10=0.7
3/20=0.15
4/25=0.16
9/50=0.18
1/8=0.125
Step-by-step explanation:
1/2=0.5
3/4=0.75
7/10=0.7
3/20=0.15
4/25=0.16
9/50=0.18
1/8=0.125
Answer:
<h3>
B(10, 6)</h3>
Step-by-step explanation:
If P is midpoint of AB and: 
then:
The number is near zero, so it is classified as "low." Its sign is negative, so it is a ...
-low negative correlation
