The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.964 g and a standard deviation of 0.302 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 30 cigarettes with a mean nicotine amount of 0.854 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 30 cigarettes with a mean of 0.854 g or less. P(M<0.854 g) = Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Assume that any probability below 5% is sufficient to conclude that the new product really have lower nicotine. Based on the result above, is it valid to claim that the amount of nicotine is lower? Yes. The probability of this data is unlikely to have occurred by chance alone. No. The probability of obtaining this data is high enough to have been a chance occurrence. C

At one college, GPA's are normally distributed with a mean of 3 and a standard deviation of 0.6. Fill in the blanks, 99.7% of GPAs are between ____ and ____.   A. 1.8, 4.2   B. 3, 3.6   C. 2.4, 3.6   D. 1.2, 4.8

The time that college students spend studying per week have a distribution that skewed to the right with a mean of 8.4 hours and a standard deviation of 2.7 hours.  Find the probability that the mean time spent studying per week for a random sample of 40 is less than 7.5 hours.

Based on a recent​ study, the pH level of the arterial cord​ (one vessel in the umbilical​ cord) is normally distributed with mean 7.27and standard deviation of 0.11. Find the percentage of preterm infants who have the following arterial cord pH levels.   a. pH levels between 7.00 and 7.50. b. pH levels over 7.35.   a. The percentage of arterial cord pH levels that are between 7.00and 7.50 is __________​%. ​(Round to two decimal places as​ needed.)   b. The percentage of arterial cord pH levels that are over 7.35 is _______%. ​(Round to two decimal places as​ needed.)