The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and standard deviation 129 chips.​(a) What is the probability that a randomly selected bag contains between 1000 and 1500 chocolate​ chips, inclusive? ​(b) What is the probability that a randomly selected bag contains fewer than 1025 chocolate​ chips? ​(c) What proportion of bags contains more than 1200 chocolate​ chips? ​(d) What is the percentile rank of a bag that contains 1425 chocolate​ chips?

Using the normal distribution, it is found that:

a) There is a 0.947 = 94.7% probability that a randomly selected bag contains between 1000 and 1500 chocolate​ chips, inclusive.

b) There is a 0.0392 = 3.92% probability that a randomly selected bag contains fewer than 1025 chocolate​ chips.

c) 0.3446 = 34.46% of bags contains more than 1200 chocolate​ chips.

d) The bag is in the 91st percentile.

In a normal distribution with mean and standard deviation, the z-score of a measure X is given by:

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Mean of 1252 chips, thus .
• Standard deviation of 129 chips, thus .

Item a:

The probability is the p-value of Z when X = 1500 subtracted by the p-value of Z when X = 1000, thus:

X = 1500:

has a p-value of 0.9726.

X = 1000:

has a p-value of 0.0256.

Then, 0.9726 - 0.0256 = 0.947

There is a 0.947 = 94.7% probability that a randomly selected bag contains between 1000 and 1500 chocolate​ chips, inclusive.

Item b:

This probability is the p-value of Z when X = 1025, thus:

has a p-value of 0.0392.

There is a 0.0392 = 3.92% probability that a randomly selected bag contains fewer than 1025 chocolate​ chips.

Item c:

This proportion is 1 subtracted by the p-value of Z when X = 1200, thus:

has a p-value of 0.3446.

0.3446 = 34.46% of bags contains more than 1200 chocolate​ chips.

Item d:

This percentile is the p-value of Z when X = 1425, thus:

has a p-value of 0.91.

The bag is in the 91st percentile.

A similar problem is given at brainly.com/question/13680644

Final answer:

The probability and percentiles can be found using the z-score formula and then looking these z-scores up in the standard normal distribution table.

Explanation:

In this question, we're dealing with a normal distribution. For a normal distribution, probabilities can be calculated using the z-score formula which is given by Z = (X - μ) / σ, where X is the value from which you want to find the probability, μ is the mean and σ is the standard deviation.

(a) To find the probability that a randomly selected bag has between 1000 and 1500 chocolate chips, we need to find the z-scores for both 1000 and 1500 and then find the area between these two z-scores in the standard normal distribution table.  (b) To find the probability that a randomly selected bag has less than 1025 chips, we find the z-score for 1025 and find the area to the left of it in the standard normal distribution table. (c) To find the proportion of bags containing more than 1200 chips, we find the z-score for 1200 and then find the area to the right of it in the standard normal distribution table. (d) Percentile rank can be found by finding the z-score for 1425 chips, and then finding the corresponding percentile in the standard normal distribution table.

Learn more about Normal Distribution here:

brainly.com/question/34741155

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