Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 167.1-cm and a standard deviation of 0.6-cm. For shipment, 6 steel rods are bundled together. Round all answers to four decimal places if necessary.

1. What is the distribution of XX? XX ~ N( , )
2. What is the distribution of ¯xx¯? ¯xx¯ ~ N( , )
3. For a single randomly selected steel rod, find the probability that the length is between 166.9-cm and 167-cm.
4. For a bundled of 6 rods, find the probability that the average length is between 166.9-cm and 167-cm.
5. For part d), is the assumption of normal necessary?NO OR YES

Answer 1

Solution:

a)  μ = 167.1 cm and  σ = 0.6 cm

Hence, σ^{​​​​​​2} = (0.6)² = 0.36

Normal distribution has standard form as N(μ, σ^{​​​​​​2}).

The distribution of X is normal distribution with mean 167.1 cm and variance 0.36 cm^{​​​​​2}.

Hence, X ~ N(167.1, 0.36)

b) If X ~ N(μ, σ^{​​​​​​2}) then, x̄ ~ N(μ, σ^{​​​​​​2}/n)

Where, n is sample size, x̄ is sample mean.

We have, σ^{​​​​​​2} = 0.36 and n = 6

σ²/n = 0.36/6 = 0.06

The distribution of x̄ is normal with mean 167.1 cm and variance 0.06 cm^{​​​​​2}.

Hence, x̄ ~ N(167.1, 0.06)

c) We have to find P(166.9 < X < 167).

We know that if X ~ N(μ, σ^{​​​​​​2}) then

We have, μ = 167.1 cm and  σ = 0.6 cm

Using "NORM.S.DIST" function of excel we get,

P(Z < -0.1667) = 0.4338 and P(Z < -0.3333) = 0.3694

For a single randomly selected steel rod, the probability that the length is between 166.9-cm and 167-cm is 0.0644.

d) We have to find P(166.9 <  x̄ < 167).

If X ~ N(μ, σ^{​​​​​​2}) then, x̄ ~ N(μ, σ^{​​​​​​2}/n)

And if x̄ ~ N(μ, σ^{​​​​​​2}/n) then,

We have, μ = 167.1 cm and  σ = 0.6 cm and n = 6

Using "NORM.S.DIST" function of excel we get,

P(Z < -0.4082) = 0.3416 and P(Z < -0.8165) = 0.2071

For a bundled of 6 rods, the probability that the average length is between 166.9-cm and 167-cm 0.1345.

e) Yes for part (d) the assumption of normal is necessary.