Question

2. Birth weights are normally distributed with a mean of 7.6 pounds and a standard deviation of 1.23 pounds. What is the probability that a newborn weighs more than 11.3 pounds? Ans 2 3. X is binomial with n = 700 and p = .32. Use the standard normal distribution to approximate P(207 < X < 256). Ans 3 4. A population has a known variance of 22.9. If you draw random samples of size 24 and construct the sampling distribution of X, what would be the value of ox? 5. Ans 4 A random sample of 24 brand A cigarettes has a mean nicotine level of 31.6 mg and a standard deviation of 3.1 mg. Using a 98% C.I., estimate the mean nicotine level for all brand A cigarettes. Ans 5

Answer 1

Solution :

2) Let X be a random variable which represents the birth weights.

Given that, X ~ N(7.6, 1.23²)

Mean (μ) = 7.6 pounds

SD (σ) = 1.23 pounds

We have to find P(X > 11.3 pounds).

We know that if X ~ N(μ, σ²) then,

X-M Z=1 ~ N(0,1) o

X-M 11.3- ::P(X> 11.3) = P 0

$\large \therefore P(X > 11.3) = P\left (Z > \frac{11.3-7.6}{1.23} \right )$

$\large \therefore P(X > 11.3) = P\left (Z > 3.0081 \right )$

Using "pnorm" function of R we get, P(Z > 3.0081) = 0.0013

$\large \therefore P(X > 11.3) = 0.0013$

Hence, the required probability is 0.0013.