Question

Let X denote the number of flaws along a 100-m reel of magnetic tape (an integer-valued variable). Suppose Zdenote the number of flaws along a 100-m reel of magnetic tape (an integer-valued variable). Suppose has approximately a normal distribution with μ =25 and σ = 5 . Use the continuity correction to calculate the probability that the number of flaws is Between 20 and 30, inclusive.
At most 30. Less than 30.

Answer 1

Answer:

a) $P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)$

$P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)$

And we can find this probability like this:

$P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728$

b) $P(X \leq 30)= P(X$

And using the z score we got:

$P(X$

c) $P(X$

And if we use the continuity correction we got:

$P(X$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Continuity correction means that we need to add and subtract 0.5 before standardizing the value specified.

Part a

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(25,5)$  

Where $\mu=25$ and $\sigma=5$

Part a

For this case we want to find this probability:

$P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)$

And if we use the z score given by:

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

We got this:

$P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)$

And we can find this probability like this:

$P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728$

Part b

For this case we want this probability:

$P(X \leq 30)$

And if we use the continuity correction we got:

$P(X \leq 30)= P(X$

And using the z score we got:

$P(X$

Part c

For this case we want this probability:

$P(X$

And if we use the continuity correction we got:

$P(X$