Question

Suppose that the total carbohydrates in the listed cereals are approximated by a normal curve with a mean of 30.7931 and a standard deviation of 8.8858. What is the probability that a randomly selected cereal has at most 25 grams of carbohydrates? What is the probability that a randomly selected cereal has at least 45 grams of carbohydrates? What is the probability that a randomly selected Kellogg’s cereal has between 30 and 40 grams of carbohydrates? What amount of carbohydrates would separate the lowest 10% of Kellogg’s cereals? What amount of carbohydrates would separate the highest 5% of Kellogg’s cereals? Show your work or calculator syntax

Answer 1

Mean, = 30.7931 g

Standard deviation, = 8.8858 g

Let X be a random variable which denotes the amount of Carbohydrates(in grams) in Kellogg's cereal

Probability that a randomly selected cereal has at most 25 grams of carbohydrates = P(X ≤ 25)

= P{Z ≤ (25 - 30.7931)/8.8858}

= P(Z ≤ -0.652)

= 0.2572

Using Calculator, this probability can be found out using function normalcdf(-1E99, 25, 30.7931, 8.8858)

Probability that a randomly selected cereal has at least 45 grams of carbohydrates = P(X ≥ 45)

= P{Z ≥ (45 - 30.7931)/8.8858}

= P(Z ≥ 1.6)

= 0.0548

Probability that a randomly selected cereal has between 30 and 40 grams of carbohydrate = P(30 ≤ X ≤ 50)

= P{(30 - 30.7931)/8.8858 ≤ Z ≤ (40 - 30.7931)/8.8858}

= P(-0.089 ≤ Z ≤ 1.036)

= 0.3854

Corresponding to lowest 10%, the z value = -1.2845

Thus, the amount of carbohydrates that separates the lowest 10% of Kellogg's cereal = 30.7931 - 1.2845*8.8858

= 19.38 grams

Corresponding to highest 5%, the z value = 1.645

Thus, the amount of carbohydrates that separates the

highest 5% of Kellogg's cereal = 30.7931 + 1.645*8.8858

= 45.41 grams