Question

random sample of air temperatures in December is taken from n 42cities in Minnesota. Temperature data is nown to be normally distributed. The historical average temperature in Minnesota in December is -8.5 and tandard deviation O=5. Use Table 1 Calculate the expected temperature value and the standard error for the sampling distribution of the sample mean (Negative values should be indicated by a minus sign. Round "expected value to 1 decimal place and "standard error" to 4 decimal places.) Expected value Standard error What is the probability that the sample mean temperature is less than -97 (Round final answer to 4 decimal places.) Probability What is the probability that the sample mean temperature falls between -9 and -8? (Do not round Intermediate calculations, round final answer to 4 decimal places.) Probability < Prev 21 of 30 M Next >

Answer 1

a) Expected value $=\mu= -8.5$

Standard error $= \frac{\sigma }{\sqrt{n}} = \frac{5}{\sqrt{42}} = 0.7715$

Let $\bar{X}$ denotes the average temperature for n =42 cities.

b)

Here, - Normal(-8.50, 5:008) or - Normal(-8.50, 0.7715172) Now, X - -8.500 -9.000 – -8.500 P(X < –9.00) = P(X<-9.00) = P(- 0.771517 - 0.771517 -9.000 – -8.500 + P(X <-9.00) = P 0.771517 + P(XS-9.00) = P(Z < -0.648074) + P(7 < –9.00) = 0.258469 – 0.2585

c) The probability that the sample mean temperature falls between -9 and -8 is

< X - -8.50 -8.00 – -8.50 X - -8.50 P(-9.00 < X <-8.00) = P(X = -8.00) – P(X3–9.00) = P( 771517 S 0271517 ) – Pl 0 771517 S -8.00 – -8.50 -9.00 – -8.50 → P(-9.00 < X <-8.00) = P(Z < 0771617 ) – P(Z < 0721617 ) -9.00 – -8.50 0271517) → P(–9.00 < ĦS -8.00) = P(Z < 0.648074) – P(Z < -0.648074) = P(-9.00 < X <-8.00) = 0.741531 – 0.258469 = 0.483063 ~0.4831