2a) assume that adult males have footless which are normally distributed with mean 24.6 cm and standard deviation 1.05 cm. Calculate the probability that an adult male has a foot length is greater than 27 cm.
My answer - 2
2b) assume adult females have foot lengths which are normally distributed with mean 22.8 cm and standard deviation 0.89 cm. Calculate the percentage of adult female populated that has foot lengths between 22 and 25.
How would I tackle part 2b?
2c) a random sample of 10 adult females are selected. Find the probability that the sample men a is less than 22.9 cm
How would I do part c as well?
Please help
Thankyou
you can figure these out at
http://davidmlane.com/hyperstat/z_table.html
To tackle part 2b, calculating the percentage of adult female populations that have foot lengths between 22 and 25 cm, you can use the cumulative distribution function (CDF) of the normal distribution.
First, you need to calculate the z-score for each of the values using the formula:
z = (x - μ) / σ
Where:
- x is the value you want to find the probability for
- μ is the mean
- σ is the standard deviation
Calculating the z-scores for 22 and 25 cm using the given information:
For 22 cm:
z1 = (22 - 22.8) / 0.89
For 25 cm:
z2 = (25 - 22.8) / 0.89
Now, you can use a standard normal distribution table or a calculator to find the probabilities associated with each z-score.
P(22 cm ≤ x ≤ 25 cm) = P(z1 ≤ z ≤ z2)
Finally, subtracting the lower probability (P(z ≤ z1)) from the higher probability (P(z ≤ z2)) will give you the percentage of adult female population with foot lengths between 22 and 25 cm.
To tackle part 2c, finding the probability of the sample mean being less than 22.9 cm for a random sample of 10 adult females, you can use the sampling distribution of the sample mean.
First, you need to calculate the standard deviation of the sampling distribution (also known as the standard error), using the formula:
Standard Error (SE) = σ / √n
Where:
- σ is the population standard deviation
- n is the sample size
In this case, the population standard deviation is given as 0.89 cm and the sample size is 10, so:
SE = 0.89 / √10
Next, you need to calculate the z-score for the sample mean using the formula:
z = (x̄ - μ) / SE
Where:
- x̄ is the sample mean
- μ is the population mean
In this case, the population mean is not given, so we assume it to be the same as the mean for adult females (22.8 cm).
z = (22.9 - 22.8) / SE
Now, you can use a standard normal distribution table or a calculator to find the probability associated with the z-score you calculated. P(z ≤ z-score)
This probability will give you the likelihood of getting a sample mean less than 22.9 cm for a random sample of 10 adult females.