Question

QUESTION PART A: The lengths of pregnancies in a small rural village are normally distributed with a mean of 263 days and a standard deviation of 15 days.

In what range would you expect to find the middle 98% of most pregnancies?
Between ____ days and____days.

If you were to draw samples of size 50 from this population, in what range would you expect to find the middle 98% of most averages for the lengths of pregnancies in the sample?
Between ____ days and ___ days.

Enter your answers as numbers to the nearest whole days.

QUESTION PART B:

Suppose the true proportion of voters in the county who support a school levy is 0.48. Consider the sampling distribution for the proportion of supporters with sample size n = 199.

What is the mean of this distribution?

What is the standard error (i.e. the standard deviation) of this sampling distribution, rounded to three decimal places?    

Answer 1

a)

µ = 39.2      

σ = 1.9

proportion=0.9800

proportion left 0.0200is equally distributed both left and right side of normal curve

z value at0.01= ±2.326(excel formula =NORMSINV(0.02/ 2 ) )

z = ( x - µ ) / σ

so, X = z σ + µ =

X1 =-2.326*1.9+39.2=34.7799

X2 =2.326*1.9+39.2=43.6201

answer: Between __35__ days and__44__days.

b)

µ = 263        

σ = 15

n=50

proportion=0.98

proportion left 0.02is equally distributed both left and right side of normal curve  

z value at0.01= ±2.33(excel formula =NORMSINV(0.02/ 2 ) )

z = ( x - µ ) / (σ/√n)

so, X = z σ / √n + µ =

X1 =-2.33*15/ √50+263=258.0651

X2 =2.33*15/ √50+263=267.9349

between 258 days to 268 days

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B)

a) mean = 0.48

b) std error , SE = √( p(1-p)/n ) = √(0.48*0.52/199) = 0.035