Questions: 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 1125 chocolate chips? (c) What proportion of bags contains more than 1225 chocolate chips? (d) What is the percentile rank of a bag that contains 1025 chocolate chips?

Solution Steps

To find the probability that a randomly selected bag contains between 1000 and 1400 chocolate chips, we calculate:

\[ P(1000 \leq X \leq 1400) = \Phi(Z_{end}) - \Phi(Z_{start}) \]

where \( Z_{end} = \frac{1400 - 1252}{129} \approx 1.1473 \) and \( Z_{start} = \frac{1000 - 1252}{129} \approx -1.9535 \).

Using the cumulative distribution function \( \Phi \):

\[ P(1000 \leq X \leq 1400) = \Phi(1.1473) - \Phi(-1.9535) \approx 0.849 \]

Next, we calculate the probability that a randomly selected bag contains fewer than 1125 chocolate chips:

\[ P(X < 1125) = \Phi(Z_{end}) - \Phi(-\infty) \]

where \( Z_{end} = \frac{1125 - 1252}{129} \approx -0.9845 \).

Thus, we have:

\[ P(X < 1125) = \Phi(-0.9845) - \Phi(-\infty) \approx 0.1624 \]

Finally, we find the proportion of bags that contain more than 1225 chocolate chips:

\[ P(X > 1225) = \Phi(\infty) - \Phi(Z_{start}) \]

where \( Z_{start} = \frac{1225 - 1252}{129} \approx -0.2093 \).

Therefore, we calculate:

\[ P(X > 1225) = \Phi(\infty) - \Phi(-0.2093) \approx 0.5829 \]

Thus, the proportion of bags containing more than 1225 chips is:

\[ P(X > 1225) = 1 - P(X \leq 1225) \approx 0.4171 \]

Final Answer