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The next two questions (7 and 8) refer to the following: The weight of bags of...

The next two questions (7 and 8) refer to the following:

The weight of bags of organic fertilizer is normally distributed with a mean of 60 pounds and a standard deviation of 2.5 pounds.

7. What is the probability that a random sample of 33 bags of organic fertilizer has a total weight between 1963.5 and 1996.5 pounds?

8. If we take a random sample of 9 bags of organic fertilizer, there is a 75% chance that their mean weight will be less than what value? Keep 4 decimal places in intermediate calculations and report your final answer to 4 decimal places.

The next two questions (8 and 9) refer to the following:

Question 10 and 11

Suppose that 40% of students at a university drive to campus.

10. If we randomly select 100 students from this university, what is the approximate probability that less than 35% of them drive to campus?

Keep 6 decimal places in intermediate calculations and report your final answer to 4 decimal places.

11. If we randomly select 100 students from this university, what is the approximate probability that more than 50 of them drive to campus?

Keep 6 decimal places in intermediate calculations and report your final answer to 4 decimal places.

12. Suppose that IQs of adult Canadians follow a normal distribution with standard deviation 15. A random sample of 30 adult Canadians has a mean IQ of 112.

We would like to construct a 97% confidence interval for the true mean IQ of all adult Canadians. What is the critical value z* to be used in the interval? (You do not need to calculate the calculate the confidence interval. Simply find z*. Input a positive number since we always use the positive z* value when calculating confidence intervals.)

Report your answer to 2 decimal places.

Solutions

Expert Solution

7) mean of total weight=60*33=1980

std dev of total weight=√33*2.5=14.36

we need to calculate probability for ,                                      
P (   1963.5   < X <   1996.5   )                      
=P( (1963.5-1980)/14.36 < (X-µ)/σ < (1996.5-1980)/14.36)                                      
                                      
P (    -1.149   < Z <    1.149   )                       
= P ( Z <    1.149   ) - P ( Z <   -1.149   ) =    0.8747   -    0.1253   =    0.7494   (answer)

8)

µ =    60                              
σ =    2.5                              
n=   9                              
proportion=   0.7500                              
                                  
Z value at    0.75   =   0.674   (excel formula =NORMSINV(   0.75   ) )          
z=(x-µ)/(σ/√n)                                  
so, X=z * σ/√n +µ=   0.674   *   2.5   / √    9   +   60   =   60.5621(answer)

milcah answered 1 year ago
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