Question

A distribution of values is normal with a mean of 165.4 and a standard deviation of 22.7 . Find the probability that a randomly selected value is less than 217.6. P(X<217,6)=_________ Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. A distribution of values is normal with a mean of 150 and a standard deviation of 4 . Find the interval containing the middle-most 32% of scores:_________ Enter your answer using interval notation. Example: [2.1,5.6) Your numbers should be accurate to 1 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

To find the probability that a randomly selected value is less than 217.6, calculate the z-score and look up the area under the normal curve. The probability is approximately 0.9890. For the interval containing the middle-most 32% of scores, use z-scores corresponding to the lower and upper quartiles.

Step-by-step explanation:

To find the probability that a randomly selected value is less than 217.6, we need to calculate the z-score for that value and then look up the corresponding area under the normal curve.

The formula to calculate the z-score is: z = (x - mean) / standard deviation

In this case, the mean is 165.4 and the standard deviation is 22.7. So, the z-score is: z = (217.6 - 165.4) / 22.7 = 2.3

Using a z-table or a calculator, we can find that the area to the left of a z-score of 2.3 is approximately 0.9890. Therefore, the probability that a randomly selected value is less than 217.6 is approximately 0.9890.

To find the interval containing the middle-most 32% of scores, we need to find the z-scores that correspond to the lower and upper percentiles.

Since the distribution is normal, we can use a z-table to find the z-scores. The z-score that corresponds to the lower quartile (16th percentile) is approximately -0.9944, and the z-score that corresponds to the upper quartile (84th percentile) is approximately 0.9944.

Therefore, the interval containing the middle-most 32% of scores is approximately [-0.9944, 0.9944].