Question

Assume that​ women's heights are normally distributed with a mean given by `mu = 62.4` in​, and a standard deviation given by `sigma = 2.8` in. ​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in? ​

​

Answer 1

`0.9895`

Explanation:

From the mean to the value of interest: `63.0-62.4 = 0.6` inches

So the count of standard deviations beyond the
mean of 63 is: `0.6/2.8sigma`

So the all encompassing probability from the left for `mu+6/26 sigma=0.9895`

Used 'Elementary Statistics Tables by Henry R Neave

Answer 2

Using a `z`-table: `58.3%`.
Using software: `58.48379%`.

Explanation:

The question is asking, if `X` is Normal with mean `mu=62.4` and standard deviation `sigma = 2.8`, then what is `"P"(X<63)?`

We can "translate" this to the Standard Normal distribution `Z`, which has `mu=0` and `sigma = 1`, by using the equivalence

`"P"(X < x) = "P"(Z< (x-mu)/sigma)`

`Z` is easier because there are `z`-tables that, for many different values of `z`, give the corresponding probability of `"P"(Z < z).`

We get:

`"P"(X < 63) = "P"(Z< (63-62.4)/2.8)`

`color(white)("P"(X < 63)) = "P"(Z< 0.6/2.8)`

`color(white)("P"(X < 63)) ~~ "P"(Z< 0.2143)`

Now all we do is look up this probability in a `z`-table. A table will give

`"P"(Z<0.21)=0.5832`

and

`"P"(Z<0.22)=0.5871`

Since 0.2143 is closer to 0.21 than to 0.22, we use the first one to get our approximation.

Alternatively, we can use a web app (or statistical software) to compute a more accurate answer of 0.5848379.