The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally distributed with mean 12521252 and standard deviation 129129 chips. ​(a) What is the probability that a randomly selected bag contains between 10001000 and 15001500 chocolate​ chips? ​(b) What is the probability that a randomly selected bag contains fewer than 10251025 chocolate​ chips? ​(c) What proportion of bags contains more than 11751175 chocolate​ chips? ​(d) What is the percentile rank of a bag that contains 10501050 chocolate​ chips?

Answer:A) 0.947; B) 0.0392; C) 0.7257; D) 6thStep-by-step explanation:For part A,We find the z-score for both of these values and subtract them; this will give us the area under the curve between the scores, which is the same as the probability between them.[tex]z=\frac{X-\mu}{\sigma}\\\\z=\frac{1000-1252}{129}\text{ and } z=\frac{1500-1252}{129}\\\\z=\frac{-252}{129}\text{ and } z=\frac{248}{129}\\\\z=-1.95\text{ and }z=1.92[/tex]Using a z-table, we see that the area to the right of z = -1.95 is 0.0256.  The area to the right of z = 1.92 is 0.9726.  This means the area between them is0.9726 - 0.0256 = 0.947.For part B,To find the probability that fewer than 1025 chips are in the bag, we find the z-score:[tex]z=\frac{X-\mu}{\sigma}=\frac{1025-1252}{129}=\frac{-227}{129}\\\\=-1.76[/tex]Looking this number up in the z-table, we find the area under the curve to the left of, or less than, this is 0.0392.For part C,Once we find the z-score for the value 1175, the z-table chart will give us the area under the curve less than this.  To find the proportion greater than this, we subtract from 1:[tex]z=\frac{X-\mu}{\sigma}=\frac{1175-1252}{129}=\frac{-77}{129}=-0.60[/tex]In the z-table, we see that the area under the curve less than this is 0.2743.  This means that the area greater than this is 1-0.2743 = 0.7257.For part D,We again find the area under the curve less than this.  This tells us the proportion of values that will be less than this; this will tell us the percentile value for this.[tex]z=\frac{1050-1252}{129}=\frac{-202}{129}=-1.57[/tex]In the z-table, we see the area to the right of this is 0.0582.  This means that 5.82% of values are less than this; this means the value is the 5.82 percentile, which rounds to the 6th percentile.