A statistics student would like to estimate the average age of students at her college. she would like to be 85% confident that her estimate includes the actual college population mean

It means confidence level is 0.85.

After taking survey the estimate of population standard deviation is found as 4.5 years. The given margin of error is 0.5 years

The sample size is obtained using formula

n= $(\frac{z_{\alpha/2} std}{ME} )^{2}$

Where std = standard deviation

alpha = 1 -c = 1-0.85 = 0.15

z(α/2) = z(0.15/2) = z(0.075)

It is the critical value of z such that area above z and below -z is 0.075

Use z score table to find the probability exactly or very close to 0.075 and note the corresponding z score value.

In z score table there is no accurate 0.075 probability value. The close value is 0.0749 which corresponds to z score -1.44

For calculation we take positive z score value.

So we get z(α/2) =1.44

Using all the given values into formula

n = [(1.44 * 4.5) / 0.5]^2

n = 167.96

Rounding it to nearest integer we get n=168

The number of people need to surveyed is n=168

7 0

2 years ago

I need first one and the second one please! I need it tomorrow!

ArbitrLikvidat [17]

I have a total of 11 pencils = total outcome
a- since we have 6 Red, P(Red)=6/11
b- P(Black) = 3/11
c- P(Green) = 1/11
For the dice:
Total outcome 6 so :
P(3) =1/6
P(1) =1/6
P(less than5) means P( of 1 OR 2 OR 3 Or 4)
Since we have "OR" it means we can add the P(probabilities)====>
P( of 1 OR 2 OR 3 Or 4)=1/6+1/6+1/6+1/6=
= 4/6 = 2/3

5 0

2 years ago