The library checks out an average of 322 books per day, with a standard deviation of 81 books per day. (The number of book check-outs per day is approximately normally distributed)Find the probability that (to 4 dec pl)a) on one day the number of books checked out is less than 200.b) the  mean number of books checked out per day for 32 days will be between 300 and 333.

a) To find the probability that on one day the number of books checked out is less than 200, we first need to standardize the value 200 using the z-score formula:

Z = (X - μ) / σ

where X is the value we are interested in (200 in this case), μ is the mean (322), and σ is the standard deviation (81).

Z = (200 - 322) / 81 = -1.51 (rounded to two decimal places)

We then look up this z-score in a standard normal distribution table or use a calculator with a normal distribution function to find the probability. The probability that Z is less than -1.51 is 0.0655 or 6.55%.

b) To find the probability that the mean number of books checked out per day for 32 days will be between 300 and 333, we first need to find the standard deviation of the mean. The standard deviation of the mean is the standard deviation divided by the square root of the number of observations (σ / √n), which in this case is 81 / √32 = 14.32 (rounded to two decimal places).

We then standardize the values 300 and 333:

Z1 = (300 - 322) / 14.32 = -1.54 (rounded to two decimal places) Z2 = (333 - 322) / 14.32 = 0.77 (rounded to two decimal places)

We then find the probabilities associated with these z-scores. The probability that Z is less than -1.54 is 0.0618 or 6.18%, and the probability that Z is less than 0.77 is 0.7794 or 77.94%.

The probability that the mean number of books checked out per day for 32 days will be between 300 and 333 is the difference between these probabilities: 0.7794 - 0.0618 = 0.7176 or 71.76%.