Question

Candles A company produces candles. Machine 1 makes candles with a mean

length of $15$cm and a standard deviation of $0.15$cm. Machine 2 makes candles with a

mean length of $15$cm and a standard deviation of $0.10$cm. A random sample of $49$

candles is taken from each machine. Let x ̄1−x ̄2 be the

difference (Machine 1 – Machine 2) in the sample mean length of candles. Describe the

shape, center, and variability of the sampling distribution of x ̄1−x ̄2.

Short answer

Shape: Shape is approximately normal

Center: $\mu$_{x ̄1−x ̄2}=$0$

Variability:$\sigma$_{x ̄1−x ̄2}=$0.026$cm

Step-by-step solution

Step 1:Given Information

We have been given that,

Machine 1 :

candles with a mean length,$x_1$= $15$cm

candles with a standard deviation,${\sigma }_1$= $0.15$cm

random sample of candles (n₁)=$49$

Machine 2:

candles with a mean length,$x_2$= $15$cm

candles with a standard deviation,${\sigma }_2$= $0.10$cm

random sample of candles (n₂)=$49$

Step 2:Explanation

n₁ >$30$ and n₂ >$30$

So, x ̄1−x ̄2 has approximately normal shape.

${\mu }_{x1-x2}=x_1-x_2$=$15-15=0$

So,center =$0$

Variability:

localid="1654269722311" ${\sigma }_{x_1-x_2}=\sqrt{\frac{{{\sigma }_1}^2}{n_1}+\frac{{{\sigma }_2}^2}{n_2}}$=localid="1654269775856" $\sqrt{\frac{0.{15}^2}{49}+\frac{0.{10}^2}{49}}$=localid="1654269783223" $0.026$cm