SOLUTION: 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 selecte

Question 1183555: 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:

more than KES 800000?
between KES 500000 and KES 700000?
less than KES 600000?
between KES 300000 and KES 600000?

Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
.
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

(a)  more than KES 800000?

     P = 0.0013.    ANSWER



(b)  between KES 500000 and KES 700000?

     P = 0.4772.    ANSWER



(c) less than KES 600000?

    P = 0.8413.     ANSWER



(d) between KES 300000 and KES 600000?

    P = 0.8186.    ANSWER

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.

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