Assume that women's heights are normally distributed with a mean given by u=64.6 in
and a standard deviation given by 0=2.8in.
a) If 1 woman is randomly selected, find the probability that her height is less than 65in.
b) If 49 women are randomly selected, find the probability that they have a mean height less than 65 in.
Assume that women's heights are normally distributed with a mean given by mu equals 62.3 inμ=62.3 in, and a standard deviation given by sigma equals 2.7 inσ=2.7 in.
(a) If 1 woman is randomly selected, find the probability that her height is less than 6363 in.
(b) If 4343 women are randomly selected, find the probability that they have a mean height less than 6363 in.
Assume that women's heights are normally distributed with a mean given by mu equals inμ=62.6 and a standard deviation given by sigma equals inσ=2.4 in.
(a) If 1 woman is randomly selected, find the probability that her height is less than 63 in.
(b) If 36 women are randomly selected, find the probability that they have a mean height less than 63 in.
(a) The probability is approximately
nothing.
(Round to four decimal places as needed.)
To answer parts a) and b), we need to use the normal distribution and its associated z-scores. To find probabilities under the normal distribution, we can convert the values of interest to z-scores and then use a table or a calculator.
a) To find the probability that a randomly selected woman's height is less than 65 inches, we need to calculate the corresponding z-score and then find the area to the left of that z-score in the standard normal distribution.
The z-score is calculated using the formula:
z = (x - μ) / σ
Given:
x = 65 inches
μ (mean) = 64.6 inches
σ (standard deviation) = 2.8 inches
Plugging the values into the formula, we get:
z = (65 - 64.6) / 2.8
z = 0.4 / 2.8
z = 0.1429
Now, we can use a z-table or a calculator to find the probability of z ≤ 0.1429. For example, using a z-table, we can find that the probability corresponding to a z-score of 0.1429 is approximately 0.5582.
Therefore, the probability that a randomly selected woman's height is less than 65 inches is approximately 0.5582.
b) To find the probability that 49 randomly selected women have a mean height less than 65 inches, we need to use the concept of the distribution of sample means. The mean of the sampling distribution of the sample mean (also called the central limit theorem) will be the same as the population mean (μ = 64.6 inches). However, the standard deviation of the sampling distribution, also known as the standard error, will be the population standard deviation divided by the square root of the sample size:
Standard error (SE) = σ / √n
SE = 2.8 / √49
SE = 2.8 / 7
SE = 0.4
We can now use the calculated standard error (SE) to find the z-score for the sample mean of 65 inches:
z = (x - μ) / SE
z = (65 - 64.6) / 0.4
z = 0.4 / 0.4
z = 1
Using a z-table or a calculator, we can find that the probability of having a z-score less than 1 is approximately 0.8413.
Therefore, the probability that 49 randomly selected women have a mean height less than 65 inches is approximately 0.8413.
a.
z = (65-64.6)/(2.8/sqrt(1))
z = 0.14
b.
z = (65-64.6)/(2.8/sqrt(49))
z = 1
Assume that women's heights are normally distributed with a mean given by mu equals 64.9 inμ=64.9 in, and a standard deviation given by sigma equals 2.7 inσ=2.7 in. Complete parts a and b.
a. If 1 woman is randomly selected, find the probability that her height is between 64.264.2 in and 65.265.2 in.
The probability is approximately
0.14650.1465. (Round to four decimal places as needed.)
b. If 6 women are randomly selected, find the probability that they have a mean height between 64.264.2 in and 65.265.2 in.
The probability is approximately
great webpage for your problem, just plug in the values that you have
http://davidmlane.com/hyperstat/z_table.html