Question

The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally...

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 1400 chocolate​ chips, inclusive? ​(b) What is the probability that a randomly selected bag contains fewer than 1100 chocolate​ chips? ​(c) What proportion of bags contains more than 1200 chocolate​ chips? ​(d) What is the percentile rank of a bag that contains 1475 chocolate​ chips? ​(a) The probability that a randomly selected bag contains between 1000 and 1400 chocolate​ chips, inclusive, is nothing. ​(Round to four decimal places as​ needed.) ​(b) The probability that a randomly selected bag contains fewer than 1100 chocolate chips is nothing. ​(Round to four decimal places as​ needed.) ​(c) The proportion of bags that contains more than 1200 chocolate chips is nothing. ​(Round to four decimal places as​ needed.) ​(d) A bag that contains 1475 chocolate chips is in the nothingth percentile. ​(Round to the nearest integer as​ needed.)

Mean M = 1252

Standard deviation s = 129

Part a)

P(1000 ≤ x ≤ 1400) = P( x≤ 1400) - P( x≤ 1000)

P(x ≤1400) = P(z ≤ (1400-1252)/129)

P(x ≤ 1400) = P(z ≤ 1.15) = 0.8749

P(x ≤ 1000) = P( z ≤ (1000-1252)/129)

P(x ≤ 1000) = P(z ≤ -1.95) = 0.0256

P(1000 ≤ x ≤ 1400) = 0.8749 -0.0256 = 0.8493

Part b)

P(x < 1100) = P(z < (1100-1252)/129)

P(x < 1000) = P(z < -1.18) = 0.1190

Part c)

P( x> 1200) = 1 - P(x < 1200)

P(x < 1175) = P(z < (1200-1252)/129)

P(x < 1175) = P(z < -0.40) = 0.3446

P(x > 1175) = 1 - 0.3446

P( x > 1175) = 0.6554

Part d)

We know that P(X < 1475) = 0.9581 = 0.96

The percentile rank is: 0.96*100

Percentile rank = 96%

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