The heights of 18 year old men are approximately normally distributed, with a mean of 67 inches and a standard deviation of 5 inches. What is the probability an 18 year old man selected at random is between 66 and 68 inches? Use FOUR decimal places.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores.
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(b) If a random sample of twenty-one 18-year-old men is selected, what is the probability that the mean height x is between 69 and 71 inches? (Round your answer to four decimal places.)
To find the probability that an 18-year-old man selected at random has a height between 66 and 68 inches, we can use the properties of a normal distribution.
The first step is to standardize the values of 66 and 68 using the standard normal distribution formula:
Z = (X - μ) / σ
Where:
Z is the standard score
X is the value we want to standardize (in this case, 66 and 68)
μ is the mean of the distribution (67)
σ is the standard deviation of the distribution (5)
For X = 66:
Z1 = (66 - 67) / 5 = -0.2
For X = 68:
Z2 = (68 - 67) / 5 = 0.2
Next, we need to find the area under the normal curve between these two Z-scores (Z1 and Z2). We can do this by looking up the cumulative probability in a standard normal distribution table.
Using the table or a calculator, the cumulative probability for Z = -0.2 is 0.4207, and the cumulative probability for Z = 0.2 is 0.5793.
The probability that an 18-year-old man selected at random has a height between 66 and 68 inches is equal to the difference between these two cumulative probabilities:
P(66 ≤ X ≤ 68) = P(Z1 ≤ Z ≤ Z2) = P(Z ≤ 0.2) - P(Z ≤ -0.2) = 0.5793 - 0.4207 = 0.1586
Therefore, the probability that an 18-year-old man selected at random is between 66 and 68 inches is approximately 0.1586 when rounded to four decimal places.