Final answer:
a) The variability in the mean number of chocolate chips in random samples of 22 chocolate chip cookies is approximately 0.93.
b) The probability that a random sample of 22 cookies will have an average less than 14.77 chocolate chips is almost 0.
c) The sample average of 14.77 suggests that the company's claim of an average of 20 chocolate chips per cookie may be inaccurate.
Step-by-step explanation:
a) The variability in the mean number of chocolate chips in random samples of 22 chocolate chip cookies can be estimated using the standard error of the mean (SEM) formula:
SEM = (standard deviation) / sqrt(sample size)
SEM = 4.37 / sqrt(22) ≈ 0.93
Therefore, the variability in the mean number of chocolate chips is approximately 0.93.
b) To find the probability that a random sample of 22 cookies will have an average less than 14.77 chocolate chips, we can calculate the z-score and use it to find the corresponding probability using a standard normal distribution table. The z-score formula is:
z = (sample mean - population mean) / (standard deviation / sqrt(sample size))
z = (14.77 - 20) / (4.37 / sqrt(22)) ≈ -4.24
Using a standard normal distribution table, the probability of obtaining a z-score less than -4.24 is very close to 0.
Therefore, the probability that a random sample of 22 cookies will have an average less than 14.77 chocolate chips is almost 0.
c) Since the probability of obtaining an average less than 14.77 is almost 0, we can conclude that the sample average of 14.77 is significantly lower than the claim on the packaging.
This suggests that the company's claim of an average of 20 chocolate chips per cookie may be inaccurate.