Question

For a standard normal distribution, find the approximate value of P (z less-than-or-equal-to 0.42). Use the portion of the standard normal table below to help answer the question.
For a standard normal distribution, find the approximate value of LaTeX: P\left(z\ge-1.25\right)P ( z ≥ − 1.25 ). Use the portion of the standard normal table below to help answer the question.

For a standard normal distribution, find the approximate value of P (negative 0.78 less-than-or-equal-to z less-than-or-equal-to 1.16). Use the portions of the standard normal table below to help answer the question.

The life of a manufacturer's compact fluorescent light bulbs is normal, with mean 12,000 hours and standard deviation 2,000 hours. Caitlynn wants to find the probability that a light bulb she purchased from this manufacturer will last at least 14,500 hours. What is the probability that the light bulb she purchases from this manufacturer will last greater than or equal to 14,500 hours? Use the portion of the standard normal table below to help answer the question.

The life spans of a computer manufacturer’s hard drives are normally distributed, with a mean of 3.5 years and a standard deviation of 0.75 years. What is the probability of a randomly selected hard drive from the company lasting between 2.25 years and 3.25 years? Use the portion of the standard normal table below to help answer the question.

Answer 1

Answer:

a) 66%

b) 89%

c) 66%

d) 11%

e) 40%

Step-by-step explanation:

a) From the standard normal distribution table:

P(z ≤ 0.42) = 0.6628 ≅ 66%

b) From the standard normal distribution table:

P(z ≥ -1.25) = 1 - P(z < -1.25) = 1 - 0.1056 = 0.8944 ≈ 89%

c) From the standard normal distribution table:

P(-0.78 ≤ z ≤ 1.16) = P(z < 1.16) - P(z < -0.78) = 0.8770 - 0.2177 = 0.6593 ≈ 66%

d) Given that:

mean (μ) = 120000 hours and standard deviation (σ) = 2000 hours.

The z score is given by the equation:

$z=\frac{x-\mu}{\sigma}$

For 14500 hours, the z score is:

$z=\frac{x-\mu}{\sigma}=\frac{14500-12000}{2000} =1.25$

P(X ≥ 14500) = P(z ≥ 1.25) = 1 - P(z < 1.25) = 1 - 0.8944 = 0.1056 ≈ 11%

e) Given that:

mean (μ) = 3.5 years and standard deviation (σ) = 0.75 years.

The z score is given by the equation:

$z=\frac{x-\mu}{\sigma}$

For 2.25 years, the z score is:

$z=\frac{x-\mu}{\sigma}=\frac{2.25-3.5}{0.75} =-1.67$

For 3.25 years, the z score is:

$z=\frac{x-\mu}{\sigma}=\frac{3.25-3.5}{0.75} =-0.33$

P(2.25 < X < 3.25) = P(-1.67 < z < -0.13) = P(z < -0.13) - P(z < -1.67) = 0.4483 - 0.0475 = 0.4008 ≈ 40%

Answer 2

The anwer is C. if *Edgenuit.y* brought you here

C. 66%