Standard deviation?

Question

the average life ofa headlamp bulb is 2,500 and the standard deviation is 225 hrs.Assuming the bulb life has a normal distribution what percentage of bulbs would need replacing;
a. before 1825hrs
b. before 2,050hrs
c before 2,900hrs
d before 3,175hrs

Answer 1

P(X < 1,825.0) = P((x - 2,500) / 225.0) < (1,825.0 - 2,500) / 225.0) = P(Z < -3.00) = 0.0013
P(X < 2,050.0) = P((x - 2,500) / 225.0) < (2,050.0 - 2,500) / 225.0) = P(Z < -2.00) = 0.0228
P(X < 2,900.0) = P((x - 2,500) / 225.0) < (2,900.0 - 2,500) / 225.0) = P(Z < 1.78) = 0.9623

P(X < 3,175.0) = P((x - 2,500) / 225.0) < (3,175.0 - 2,500) / 225.0) = P(Z < 3.00) = 0.9987

Answer 2

Just convert all these numbers to Standard Normal Distribution (centered at zero and STD=1) by subtracting the average and then dividing by the STD.

Then you can look up the answers in any old distribution table.

Answer 3

1825 hours is 3 standard deviations below the mean
2050 is 2 below the mean
2900 is 1.78 above the mean
3175 is 3 above the mean

Using a table of the cumulative normal distribution, (or the NORMDIST function in Excel) find that
: x = -3 sd --> p = .00135
: x = -2 sd --> p = 0.0228
: x = 1.78 sd --> p = .962
: x = 3 sd --> p = .999
Therefore you expect .14% of the bulbs to fail by 1825 hours, 2.3% by 2050 hours, 96.2% by 2900 and 99.9% by 3175 hours.

Answer 4

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Answer 5

a 1825 is 3SD below the mean so 0.15% of values are below this level

b 2,025 is 2SD blow the mean so 2.3% are below this value

c 2,900 is 1.7SD above the mean. The value for 2SD is 2.3% of values are above this level, so 97.7% (approx) would need replacing. (my table only gives values for 1, 2, 3 SD etc)

d 3,175 is 3SD above the mean and 98.85% would need replacing before this time.

Answer 6

Use the tables.