Question

i need a break down to the simplist level possible

In a recent year, the ACT scores on the math portion were distributed normal poslation mean of 21.1 and a population standard deviation of 53. Find the pro randomly selected student has a score between 19 and 24 7 The time spent waiting for a kidney transplant is distributed normally with mean of 1574 and a standard deviation of 212.5 days. What waiting time represents the first quarte The mean annual salary for parole officers is $50,830. A random sample of soparole s selected fa $8520, what is the probability that the sample mean annual salary is less than $50,000? in a survey of 814 U.S. adults, 375 say the economy is the most important is facing the country today. Find the point estimates for panda. Develop a 95% confidence interval for 5. The mean amount that US adults spend weekly on food is $151 with a standard deviation of $49. Random samples of sures are drawn from this population and the mean of each amples determined. What is the probability that the mean amount spend on food for a certain sample is more than $160? 6 You wish to estimate with 95% confidence the population proportion of US adults who they should be saving more money. Your estimate must be accurate within of the population. No preliminary estimate is available. Find the minimum sample s eeded

Answer 1

1)

µ =    21.1                                  
σ =    5.3                                  
we need to calculate probability for ,                                      
P (   19   < X <   24   )                      
=P( (19-21.1)/5.3 < (X-µ)/σ < (24-21.1)/5.3 )                                      
                                      
P (    -0.396   < Z <    0.547   )                       
= P ( Z <    0.547   ) - P ( Z <   -0.396   ) =    0.7079   -    0.3460   =    0.3619   (answer)

2)

µ=   1674                  
σ =    212.5                  
proportion=   0.2500                  
                      
Z value at    0.25   =   -0.674   (excel formula =NORMSINV(   0.25   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.674   *   212.5   +   1674  
X   =   1530.6709   (answer)          

3)

Z =   (X - µ )/(σ/√n) = (   50000   -   50830.00   ) / (   8520.000   / √   50   ) =   -0.69  
                                          
P(X ≤   50000   ) = P(Z ≤   -0.689   ) =   0.2455                       (answer)

4)

Number of Items of Interest,   x =   375
Sample Size,   n =    814
      
Sample Proportion , p̂ = x/n =    0.4607

q = 1 - p=1-0.4607 = 0.5393

Level of Significance,   α =    0.05
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0175          
margin of error , E = Z*SE =    1.960   *   0.0175   =   0.0342
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.461   -   0.0342   =   0.4264
Interval Upper Limit = p̂ + E =   0.461   +   0.0342   =   0.4949
                  
95%   confidence interval is (   0.4264   < p <    0.4949   )