Question

The inside diameter of a randomly selected piston ring is a random variable with mean value 14 cm and standard deviation 0.05 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.) (a) Calculate P(13.99 ≤ X ≤ 14.01) when n = 16. P(13.99 ≤ X ≤ 14.01) = 0.4207 Incorrect: Your answer is incorrect. (b) How likely is it that the sample mean diameter exceeds 14.01 when n = 25? P(X ≥ 14.01) = 0.4207 Incorrect: Your answer is incorrect. Y

Answer 1

Answer:

a) 0.5762

b) 0.1587

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 14 cm

Standard Deviation, σ = 0.05 cm

We are given that the distribution of diameter is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(13.99 ≤ X ≤ 14.01 when n = 16)

Standard error due to sampling =

$=\dfrac{\sigma}{\sqt{n}} = \dfrac{0.05}{\sqrt{16}} = 0.0125$

$P(13.99 \leq x \leq 14.01) = P(\displaystyle\frac{13.99 - 14}{0.0125} \leq z \leq \displaystyle\frac{14.01-14}{0.0125}) = P(-0.8 \leq z \leq 0.8)\\\\= P(z \leq 0.8) - P(z < -0.8)\\= 0.7881 - 0.2119 = 0.5762$

b) P(sample mean diameter exceeds 14.01 when n = 25)

Standard error due to sampling =

$=\dfrac{\sigma}{\sqt{n}} = \dfrac{0.05}{\sqrt{25}} = 0.01$

$P( x > 14.01) = P( z > \displaystyle\frac{14.01 - 14}{0.01}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,  

$P(x > 14.01) = 1 - 0.8413 = 0.1587$