Question

A. The lengths of pregnancies in a small rural village are normally distributed with a mean of 266 days and a standard deviation of 14 days.
In what range would you expect to find the middle 98% of most pregnancies?
Between_____ and___ .
If you were to draw samples of size 35 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_________ and __________.
b. Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the population of potential clientele have head breadths that are normally distributed with a mean of 6.9-in and a standard deviation of 0.9-in.
In what range would you expect to find the middle 95% of most head breadths?
Between ____________and ___________.
If you were to draw samples of size 45 from this population, in what range would you expect to find the middle 95% of most averages for the breadths of male heads in the sample?
Between____ and____ .
c. The lengths of pregnancies in a small rural village are normally distributed with a mean of 265.3 days and a standard deviation of 15.2 days.
In what range would you expect to find the middle 50% of most pregnancies?
Between ____and____ .
d. The lengths of pregnancies in a small rural village are normally distributed with a mean of 265 days and a standard deviation of 16 days.
In what range would you expect to find the middle 68% of most pregnancies?
Between _________and ___________.
If you were to draw samples of size 44 from this population, in what range would you expect to find the middle 68% of most averages for the lengths of pregnancies in the sample?
Between_____ and_____ .

Answer 1

Step-by-step explanation:

a.

mean = 266

sd = 14

cumulative probability = 0.01 so the standard score = -2.33 and 2.33 to the right and left

we find X-upper and X-lower

X-lower = 266-2.33*14 = 233.38

X-upper = 266+2.33*14 = 298.62

Between 233.38 and 298.62

we have sample size = 35

X-lower = 266-2.33*14/√35 = 260.49

X-upper = 266+2.33*14/√35 = 271.5

Between 260.49 and 271.5

b. cumulative probaility = 0.25

standard score = 1.96 to the right and left

x-lower = 6.9-1.96x0.9 = 5.14

x-upper = 6.9+1.96x0.9 = 8.66

Between 5.14 and 8.66

if sample size = 45

x-lower = 6.9-1.96*0.9/√45 = 6.64

x-upper = 6.9+1.96*0.9/√45 = 7.2

Between 6.64 and 7.2

c. standard scores would have cut off value at 0.67 and -0,67

x-lower = 265.3-0.67x15.2 = 255.12

x-upper = 265.3+0.67x15.2 = 275.48

Between 255.12 and 275.48

d. we will have critical values at 1.00 and -1.00

X-lower = 265-1x16 = 249

x-upper = 265+1x16 = 281

Between 249 and 281

with sample size = 44

x-lower = 265-1x16/√44 = 262.59

x-upper = 265+1x16/√44 = 267.41

Between 262.59 and 267.41