Question

Hi! i have an example problem with answers. Based on these answers, please solve the questions below. A sample of size 86 will be drawn from a population with a mean 90 and standard deviation 24.

(a) Find the probability that x will be less than 89. Round the answer to at least four decimal places.
The probability that × will be less than 89 is 0.3496.
(b) Find the 35th percentile of ›. Round the answer to at least two decimal places.
The 35th percentile is 88.99
A sample of size 60 will be drawn from a population with mean 31 and standard deviation 6.
(a) Find the probability that × will be between 30 and 33. Round the answer to at least four decimal places.

Answer 1

Explanation:

To solve the problem, we have the following information:

- Sample size: 60

- Mean of the population (µ): 31

- Standard deviation of the population (σ): 6

(a) To find the probability that x will be between 30 and 33, we need to use the concept of the standard normal distribution.

First, we need to convert the values 30 and 33 into z-scores using the formula:

z = (x - µ) / σ

For 30:

z = (30 - 31) / 6 = -0.1667

For 33:

z = (33 - 31) / 6 = 0.3333

Next, we use a z-table or a calculator to find the corresponding probabilities for these z-scores.

Using the z-table or a calculator, we find that the z-score of -0.1667 corresponds to a probability of approximately 0.4332, and the z-score of 0.3333 corresponds to a probability of approximately 0.6293.

To find the probability of x being between 30 and 33, we subtract the probability corresponding to the lower z-score from the probability corresponding to the higher z-score:

0.6293 - 0.4332 = 0.1961

Therefore, the probability that x will be between 30 and 33 is approximately 0.1961.