Download Quiz on Sampling Distributions

Practice Quiz 5
1) For a certain type of battery, the lifetimes of batteries under continuous use follow a Normal
distribution with mean 17 hours and standard deviation 0.8 hours.
a) What is the probability that a randomly chosen battery has a lifetime less than 16.7 hours? (Hint: this
is not a chapter 7 problem (yet) because you don’t have a random sample of batteries, just one
randomly chosen battery)
b) What are the mean and standard deviation of the sampling distribution of 𝑥̅ , the average battery
lifetime of a random sample of 60 batteries? Please use correct notation.
c) What is the probability that the average battery lifetime of a random sample of 60 batteries is less
than 16.7 hours? (Explain why you can use the Normal calculation)
2) Suppose that 47% of all adult women think they do not get enough time for themselves. An opinion
poll interviews 1025 randomly chosen women and records the sample proportion, 𝑝̂ ,who feel they don’t
get enough time for themselves.
a) What is the mean of the sampling distribution of 𝑝̂ ?
b) What is the standard deviation of the sampling distribution of 𝑝̂ ?
c) Check that the distribution of 𝑝̂ is approximately Normal.
d) Find the probability that 𝑝̂ takes on a value between 0.45 and 0.50.
Solutions
1)
a) P(X < 16.7) = P(z < -0.38) = 0.3520
b)  x = 17,  x = 0.10328
c) P( x < 16.7) = P(z < -2.90) = 0.0019, we can use the Normal calculation for TWO reasons. First, because
we were told that the population distribution is Normal. Second, because the sample size is at least 30,
and according to the Central Limit Theorem the distribution of 𝑥̅ follows approximately a Normal
distribution if the sample is large enough (at least 30). We only need ONE of these conditions to be met,
but in this case both of them were met.
2)
a) 𝜇𝑝̂ =0.47,
b) 𝜎𝑝̂ =0.01559
c) the distribution of 𝑝̂ follows a Normal distribution because both np and nq are at least 5.
1025(.47) = 482 >= 5, 1025(.53) = 543 >= 5
d) P(0.45 < 𝑝̂ < 0.50) = P(-1.28 < z < 1.92) = 0.9726 – 0.1003 = 0.8723