Question 1183555
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The average Kenyan family of four spends KES 500000 per year on food prepared at home. 
Assuming a normal distribution with a standard deviation of KES 100000 and a randomly selected 
Kenyan family of four, what is the probability that the family’s annual spending for food 
prepared at home will be:


(a) more than KES 800000?
(b) between KES 500000 and KES 700000?
(c) less than KES 600000?
(d) between KES 300000 and KES 600000?
~~~~~~~~~~~~~~



Go to online (free of charge) normal distribution probability calculator 
https://onlinestatbook.com/2/calculators/normal_dist.html



Input the given parameters of each part into the appropriate window of the calculator and get the following answers



<pre>
(a)  more than KES 800000?

     P = 0.0013.    <U>ANSWER</U>



(b)  between KES 500000 and KES 700000?

     P = 0.4772.    <U>ANSWER</U>



(c) less than KES 600000?

    P = 0.8413.     <U>ANSWER</U>



(d) between KES 300000 and KES 600000?

    P = 0.8186.    <U>ANSWER</U>
</pre>


With this technique, you do not need to think.


The ability to input data correctly into an appropriate calculator's window is totally enough to answer all these questions.



Solved.