Question

A sample size 115 will be drawn from a population of mean 48 and standard deviation 12. Find the probability that x will be greater than 45. Then find the 90th % of x

Answer 1

To find the probability that x will be greater than 45, we calculate the z-score using the formula (x - mean) / standard deviation and then look up the probability in the standard normal distribution table. The probability is 0.5962. To find the 90th percentile of x, we calculate the z-score for which the cumulative probability is 0.90 and use the formula z = (x - mean) / standard deviation to solve for x. The 90th percentile is x = 61.36.

Step-by-step explanation:

To find the probability that x will be greater than 45, we need to calculate the z-score and then use a standard normal distribution table. The formula for calculating the z-score is (x - mean) / standard deviation.

In this case, x = 45, mean = 48, and standard deviation = 12. Plugging these values into the formula, we get a z-score of -0.25. Looking up the z-score in the standard normal distribution table, we find that the probability of z > -0.25 is 0.5962.

To find the 90th percentile of x, we need to calculate the z-score for which the cumulative probability is 0.90. Plugging this probability into the standard normal distribution table, we find that the corresponding z-score is 1.28. Now, we can use the formula z = (x - mean) / standard deviation to solve for x. Plugging in the values, we get: 1.28 = (x - 48) / 12. Solving for x, we get x = 61.36.

Answer 2

The 90th percentile of x is approximately 63.36. This means that 90% of the values in the population are less than or equal to 63.36.

To find the probability that a randomly selected value from the population is greater than 45, you can use the Z-score formula and the standard normal distribution.

The Z-score is calculated as:

Z= X−μ/σ

​where:

X is the value you're interested in (45 in this case),

μ is the mean of the population (48 in this case),

σ is the standard deviation of the population (12 in this case).

So,

Z= 45−48/12​ =−0.25

Now, you can look up the probability that a Z-score is greater than -0.25 in the standard normal distribution table or use a calculator. For -0.25, the probability is about 0.5987.

Therefore, the probability that a randomly selected value from the population is greater than 45 is approximately 0.5987 or 59.87%.

Next, to find the 90th percentile of x, you'll need to find the Z-score corresponding to the 90th percentile. You can look up this value in the standard normal distribution table or use a calculator. For the 90th percentile, the Z-score is approximately 1.28.

Now, use the Z-score formula to find X (the value corresponding to the 90th percentile):

X=μ+Z×σ

X=48+1.28×12=48+15.36=63.36

Therefore, the 90th percentile of x is approximately 63.36. This means that 90% of the values in the population are less than or equal to 63.36.