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

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((19.5-25)/(5) \leq Z \leq (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<30.5)`

And using the z score we got:

`P(X<30.5) = P(Z< (30.5-25)/(5)) =P(Z<1.1) = 0.864`

c)
`P(X<30)`

And if we use the continuity correction we got:

`P(X<30-0.5) =P(Z<(29.5 -25)/(5))= P(Z<0.9) =0.816`

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 =(x-\mu)/(\sigma)`

We got this:

`P(19.5 \leq X \leq 30.5) = P((19.5-25)/(5) \leq Z \leq (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<30.5)`

And using the z score we got:

`P(X<30.5) = P(Z< (30.5-25)/(5)) =P(Z<1.1) = 0.864`

Part c

For this case we want this probability:

`P(X<30)`

And if we use the continuity correction we got:

`P(X<30-0.5) =P(Z<(29.5 -25)/(5))= P(Z<0.9) =0.816`