sampling distribution\nexamples: from previous exams\nanswer questions 25 and 26 based on this story. assume…
sampling distribution\nexamples: from previous exams\nanswer 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.\n25. what is the probability that a particular print will require more than 205 seconds to develop?\na. 0.0329\nb. 0.2266\nc. 0.75\nd. 1.84\ne. 0.7734\n26. 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?\na. $xsim n(202,4)$\nb. $overline{x}sim n(202,4/sqrt{20})$\nc. $xsim n(205,4)$\nd. $overline{x}sim n(205,4/sqrt{20})$\ne. none of the above.
Answer
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:
- B. 0.2266
- B. $\overline{X}\sim N(202,4/\sqrt{20})$