Because the sample size is large (n = 90), the central limit theorem tells us that large sample
techniques are appropriate. Accordingly,
(a) The graph of the sampling distribution is shown below:(b) From part (a), the area to the left of 11.5 is 1 – 0.0869 = 0.9131. Since the sampling distribution
is approximately normal, it is symmetric. Since 11 is the same distance to the left of the mean as
11.5 is to the right, we know that P ( < 11) = P ( > 11.5) = 0.0869. Hence, P (11 < < 11.5) =
0.9131 – 0.0869 = 0.8262. The calculator solution is normalcdf(11,11.5,11,
0.184)=0.8258.example: Over the years, the scores on the final exam for AP Calculus have been normally
distributed with a mean of 82 and a standard deviation of 6. The instructor thought that this
year’s class was quite dull and, in fact, they only averaged 79 on their final. Assuming that this
class is a random sample of 32 students from AP Calculus, what is the probability that the
average score on the final for this class is no more than 79? Do you think the instructor was
right?
solution:If this group really were typical, there is less than a 1% probability of getting an average this low
by chance alone. That seems unlikely, so we have good evidence that the instructor was correct.
(The calculator solution for this problem is normalcdf(-1000,79, 82,1.06).)Sampling Distributions of a Sample Proportion
If X is the count of successes in a sample of n trials of a binomial random variable, then the proportion of
success is given by = X /n . is what we use for the sample proportion (a statistic). The true population
proportion would then be given by p .