In: Statistics and Probability
1. assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability that a randomly selected adult has an IQ between 105 and 125.
2. 120 142 158 128 135 152
Find x̅, s, and t, construct a 90% confidence interval pf the population mean.
Solution :
Given that,
mean = = 100
standard deviation = = 15
1 ) P (105 < x < 125 )
P (105 - 100 / 15) < ( x - / ) < (125 - 100 /15)
P ( 5 / 15 < z 25 / 15 )
P (0.33 < z < 1.67)
P ( z < 1.67 ) - P ( z < 0.33)
Using z table
= 0.9525 - 0.6293
= 0.3232
Probability = 0.3232Solution:
x | x2 |
120 | 14400 |
142 | 20164 |
158 | 24964 |
128 | 16384 |
135 | 18225 |
152 | 23104 |
∑x=835 | ∑x2=117241 |
Mean ˉx=∑xn
=120+142+158+128+135+152/6
=835/6
=139.1667
Sample Standard deviation S=√∑x2-(∑x)2nn-1
=√117241-(835)26/5
=√117241-116204.1667/5
=√1036.8333/5
=√207.3667
=14.4002
Degrees of freedom = df = n - 1 = 6 - 1 = 5
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
t /2,df = t0.05,5 =2.015
Margin of error = E = t/2,df * (s /n)
= 2.015 * (14.40 / 625)
= 11.85
Margin of error = 11.85
The 90% confidence interval estimate of the population mean is,
- E < < + E
139.17 - 11.85 < < 139.17 + 11.85
127.32 < < 151.02
(127.32, 151.02 )