Question

2. The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1262 chips and a standard deviation of 118 chips
Part (a): About how many percent of the cookie bags are in between 1200 and 1300 chips.
Part (b): Determine the 30th percentile for the number of chocolate chips in a bag.
Part (c): Determine the minimum number of chocolate chips in a bag that make up the middle 95 % of bags.
Part (d): Determine the maximum number of chocolate chips in a bag that make up the middle 97 % of bags.

Answer 1

2)

Solution :

Given that ,

mean = = 1262

standard deviation = = 118

(a)

P(1200 < x < 1300) = P((1200 - 1262) / 118) < (x - ) /  < (1300 - 1262) / 118) )

= P(-0.5254 < z < 0.3220)

= P(z < 0.3220) - P(z < -0.5254)

= 0.6263 - 0.2997

= 0.3266

Probability = 0.3266

(b)

P(Z < z) = 0.30

P(Z < -0.52) = 0.30

z = -0.52

Using z-score formula,

x = z * +

x = -0.52 * 118 + 1262 = 1200.64 = 1201

30th percentile = 1201

(c)

The middle 95% has the value : -1.96 and +1.96

x = -1.96 * 118 + 1262 = 1200.64 = 1030.72 = 1031

x = 1.96 * 118 + 1262 = 1200.64 = 1493.28 = 1493

The minimum number of chocolate chips in a bag are : 1031 and 1493

(d)

The middle 95% has the value : -2.17 and +2.17

x = -2.17 * 118 + 1262 = 1200.64 = 1005.94 = 1006

x = 2.17 * 118 + 1262 = 1200.64 = 1518.06 = 1518

The minimum number of chocolate chips in a bag are : 1006 and 1518