Question

A calculator operates on two 1.5 volt batteries for a total of 3 volts. The actual

voltage of a battery is normally distributed with mean 1.5 and variance 0.045.

The tolerance in the design of the calculator is such that it will not operate

satisfactorily if the total voltage fails outside the range 2.7 to 3.3 volts. What is

the probability that the calculator will function correctly? P(0<z<1) = 0.3413​

Answer 1

Answer:

0.915

Step-by-step explanation:

A calculator operates on two 1.5 volt batteries for a total of 3 volts. The actual  voltage of a battery is normally distributed with mean 1.5 and variance 0.045.  

The tolerance in the design of the calculator is such that it will not operate  satisfactorily if the total voltage fails outside the range 2.7 to 3.3 volts. What is

the probability that the calculator will function correctly? P(0<z<1) = 0.3413​

Z score =  (Value - Mean) / Variance

Mean = 1.5V   Variance = 0.045

for 2.7 V  

Z = (2.7 - 2*1.5)/(0.045) = -0.3/0.045 = -5/3 = -1.67

95.25 % calculator after - 1.67

so 4.75 % calculator before this

for 3.3V

Z = (3.3 - 2*1.5)/(0.045) = 0.3/0.045 = 5/3 = 1.67

95.25 % calculator upto 1.67

so 4.75 % out of this

Between 2.7 V & 3.3V = 95.25 - 4.75 = 91.5 %

the probability that the calculator will function correctly = 0.915