Question

sampling distribution
examples: from previous exams
answer questions 25 and 26 based on this story. assume that the development time for a particular type of photographic printing paper when it is exposed to a light source for five seconds has a normal distribution with mean, $mu = 202$ seconds, and standard deviation, $sigma = 4$ seconds.

1. what is the probability that a particular print will require more than 205 seconds to develop?

a. 0.0329
b. 0.2266
c. 0.75
d. 1.84
e. 0.7734

1. suppose you wanted to find the probability that the mean of a random sample of 20 trials is more than 205 seconds. which distribution below would you use to standardize your probability statement?

a. $xsim n(202,4)$
b. $overline{x}sim n(202,4/sqrt{20})$
c. $xsim n(205,4)$
d. $overline{x}sim n(205,4/sqrt{20})$
e. none of the above.

Explanation:

Step1: Calculate z - score for question 25

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 205$, $\mu=202$ and $\sigma = 4$. So $z=\frac{205 - 202}{4}=\frac{3}{4}=0.75$.
We want $P(X>205)$, which is equivalent to $1 - P(X\leq205)$. Looking up the z - value of $0.75$ in the standard normal distribution table, $P(Z\leq0.75)=0.7734$. So $P(X > 205)=1 - 0.7734=0.2266$.

Step2: Determine distribution for question 26

The sampling distribution of the sample mean $\overline{X}$ for a sample of size $n$ from a normal population $X\sim N(\mu,\sigma)$ has mean $\mu_{\overline{X}}=\mu$ and standard deviation $\sigma_{\overline{X}}=\frac{\sigma}{\sqrt{n}}$. Here, $\mu = 202$, $\sigma = 4$ and $n = 20$. So $\overline{X}\sim N(202,\frac{4}{\sqrt{20}})$.

Answer:

1. B. 0.2266
2. B. $\overline{X}\sim N(202,4/\sqrt{20})$