Assume that​ women's heights are normally distributed with a mean given by mu equals 63.4 in​, and a standard deviation given by sigma equals 1.8 in. Complete parts a and b.

a. If 1 woman is randomly​ selected, find the probability that her height is between 62.7 in and 63.7 in.
The probability is approximately
nothing. ​(Round to four decimal places as​ needed.)

Did you sketch your curve to see exactly what the question is asking? Then you have to find the z-scores from both ends and use your z-score table (and remember that it reads less than) so you have to do the right most bound - the left most bound.

We will be happy to check your answer.

the height range is from 7/18 s.d. below the mean to 3/18 above the mean

a z-score table will show the portion of the population that lies in this range
... the portion is equivalent to the probability

To do this type of question, typically you will need a set tables showing standard deviation values. These were traditionally found in the back of textbooks. Many books these days no longer have these, since much more accurate on-line applets will do the same thing.

Here is one of the best: http://davidmlane.com/normal.html

Make sure you have selected "Area from a value" (the default)
Fill in your mean and SD, then click on 'between' , filling in these values

To find the probability that a woman's height is between 62.7 in and 63.7 in, we can use the z-score formula and the Z-table.

The z-score formula is given by: z = (x - μ) / σ

Where:
- x represents the given value (height)
- μ represents the population mean (63.4 in)
- σ represents the standard deviation (1.8 in)

To find the probability, we need to find the z-scores for both the lower and upper limits of the range (62.7 in and 63.7 in) and then find the difference between their cumulative probabilities.

For the lower limit of 62.7 in:
z1 = (62.7 - 63.4) / 1.8

For the upper limit of 63.7 in:
z2 = (63.7 - 63.4) / 1.8

After calculating the z-scores, we can use the Z-table (also known as the standard normal distribution table) to find the cumulative probabilities associated with z1 and z2.

Once we have the cumulative probabilities for both z1 and z2, we can subtract the lower probability from the higher probability to get the probability that her height is between 62.7 in and 63.7 in.

Here's the step-by-step calculation:

1. Calculate z1:
z1 = (62.7 - 63.4) / 1.8 = -0.3889

2. Calculate z2:
z2 = (63.7 - 63.4) / 1.8 = 0.1667

3. Use the Z-table to find the cumulative probability associated with z1:
Using the Z-table or a statistical calculator, the cumulative probability (area under the standard normal curve) for z1 of -0.3889 is approximately 0.3483.

4. Use the Z-table to find the cumulative probability associated with z2:
Using the Z-table or a statistical calculator, the cumulative probability (area under the standard normal curve) for z2 of 0.1667 is approximately 0.5662.

5. Find the probability that her height is between 62.7 in and 63.7 in:
P(62.7 in ≤ height ≤ 63.7 in) = P(z1 ≤ z ≤ z2) = P(z ≤ z2) - P(z ≤ z1) = 0.5662 - 0.3483 ≈ 0.2179

Therefore, the probability that a randomly selected woman's height is between 62.7 in and 63.7 in is approximately 0.2179.