SOLUTION: A company manufacturing light bulbs claims that an average light bulb lasts 300
days. A potential customer randomly selects 15 bulbs for testing. The sampled bulbs
last an averag
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Question 927428: A company manufacturing light bulbs claims that an average light bulb lasts 300
days. A potential customer randomly selects 15 bulbs for testing. The sampled bulbs
last an average of 290 days, with a standard deviation of 50 days. If the company’s
claim were true, what is the probability that 15 randomly selected bulbs would have
an average life of no more than 290 days?
mean = 300
P(x ≤ 290) = P( z < -10/(50/sqrt(15)) =P(z < -.7746)
Using a TI calculator 0r similarly a Casio fx-115 ES plus
P(x ≤ 290) =P(z < -.7746) = normalcdf(-100, -.7746)= .2193 0r 21.93%
Where did the -100 come from?
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
P(x ≤ 290) =P(z < -.7746) = normalcdf(-100, -.7746)
normalcdf(-100, -.7746) Gives us the area Under the Normal Curve from (-100 < z < -.7746)
z-value -100 is used as a place holder, so to speak, as it is so.... far to the left ... that basically ALL the Area
to the left of z-value -.7746 is then included. Use z-value -1000 if You wish.
................
.....Note how far z-value -3 is to the Left, for example
For the normal distribution: Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.
Area under the standard normal curve to the left of the particular z is P(z)
Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50% to the right
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