Question

1. a. Find the z value to the left of mean so that 0.025 percent of the area under the distribution curve lies to the left of it. Also report the area that was left over after the last calculation.

b. Find area between Z= 0 and Z= 1.31

           c. Find probability of P( -1.96≤ z ≤ 1.96) and P (Z=0)                  

2. The average score of a Business Studies graduate is 72 and the standard deviation is 5. The top 2 percent of the graduates receive a certificate of distinction and the next 5 percent are given a chance to apply to the business incubation center for an entrepreneurial venture.

1. What score/marks must you exceed to get a distinction and what range of marks is required for the entrepreneurial opportunity.
2. What average must you get in order to avoid failing.                  

3. The tax rates for basic necessities to be used during the COVID-19 by households items in cents is as follows:

112	120	98	55	71	35	99	124	64
150	150	55	100	132	20	70	93

1. Find the 99% and 95% confidence interval for the food tax on the selected items.
2. Which interval is larger and why?
3. If you wanted to convey the true picture of the majority of the people to the government which confidence interval would you report? Give a logical explanation.                                                                                                       

Answer 1

a) P(X≤x) =   0.025                  
                      
Z value at    0.025   =   -1.9600   (excel formula =NORMSINV(   0.025   ) )

b)     
                                      
P (    0.000   < Z <    1.310   )                       
= P ( Z <    1.310   ) - P ( Z <   0.00   ) =    0.9049   -    0.5000   =    0.4049   (answer)

c)

P (    -1.960   < Z <    1.960   )                       
= P ( Z <    1.960   ) - P ( Z <   -1.96   ) =    0.9750   -    0.0250   =    0.9500   (answer)

d) P(z=0) = 0.00

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2)

Z value at    0.98   =   2.0537   (excel formula =NORMSINV(   0.98   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   2.054   *   5   +   72  
X   =   82.27   (answer)          
marks must you exceed to get a distinction ≥ 82.27

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P(X≤x) =   0.93                  
                      
Z value at    0.93   =   1.4758   (excel formula =NORMSINV(   0.93   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.476   *   5   +   72  
X   =   79.38   (answer)      

range of marks is required for the entrepreneurial opportunity= (79.38 , 82.27 )  

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b) average must you get in order to avoid failing >79.38

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3)

a)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   38.3658
Sample Size ,   n =    17
Sample Mean,    x̅ = ΣX/n =    91.0588

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   16          
't value='   tα/2=   2.92   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   38.3658   / √   17   =   9.3051
margin of error , E=t*SE =   2.9208   *   9.3051   =   27.1781
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    91.06   -   27.178080   =   63.8807
Interval Upper Limit = x̅ + E =    91.06   -   27.178080   =   118.2369
99%   confidence interval is (   63.88   < µ <   118.24   )
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Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   16          
't value='   tα/2=   2.12   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   38.3658   / √   17   =   9.3051
margin of error , E=t*SE =   2.1199   *   9.3051   =   19.7259
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    91.06   -   19.725869   =   71.3330
Interval Upper Limit = x̅ + E =    91.06   -   19.725869   =   110.7847
95%   confidence interval is (   71.33   < µ <   110.78   )

b) 99% interval is wider