Question

A Texas Instrument calculator operating life is normally distributed with a mean of 3,850 hours, and a standard deviation of 350 hours. What is the probability that the batteries will last between 3900 and 4200 hours?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 3850

standard deviation = $\sigma$ = 350

P(3900< x <4200 ) = P[(3900-3850) / 350< (x - $\mu$ ) / $\sigma$< (4200-3850) / 350)]

= P(0.14 < Z <1 )

= P(Z < 1) - P(Z <0.14 )

Using z table   

= 0.8413 - 0.5557

probability= 0.2856