Respuesta :
Answer: The probability of a bulb lasting for at most 560 hours is 0.84
Step-by-step explanation:
Since the life of light bulbs is distributed normally, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = life of light bulbs.
µ = mean lifetime
σ = standard deviation
From the information given,
µ = 540 hours
Variance = 400
σ = √variance = √400
σ = 20
The probability of a bulb lasting for at most 560 hours is expressed as
P(x ≤ 560)
For x = 560
z = (560 - 540)/20 = 1
Looking at the normal distribution table, the probability corresponding to the z score is 0.84
Answer:
Probability of a bulb lasting for at most 560 hours = 0.8413 .
Step-by-step explanation:
We are given that the life of light bulbs is distributed normally. The variance of the lifetime is 400 and the mean lifetime of a bulb is 540 hours.
Let X = life of light bulbs
So, X ~ N([tex]\mu = 540,\sigma^{2} =20^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean and [tex]\sigma[/tex] = standard deviation
So, Probability of a bulb lasting for at most 560 hours = P(X <= 560 hours)
P(X <= 560) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{560-540}{20}[/tex] ) = P(Z <= 1) = 0.8413
Therefore, probability of a bulb lasting for at most 560 hours is 0.8413 .
