Answer:
umm it's hardd kk it but I'll help
Answer:
Approximately, the 90% confidence interval for the students' mean IQ score is between 129.045 - 130.956
Step-by-step explanation:
The formula to use to solve this question is called the Confidence Interval formula.
Confidence interval =
x ± z × ( σ/ (√n) )
Where:
x = the sample mean = 130
z = the z-value for 90% confidence = 1.645
σ = standard deviation = 7
n = sample size = 145
130 ± 1.645 × (7/√145)
130 ± 0.9562687005
130 - 0.9562687005 = 129.0437313
130 + 0.9562687005 = 130.9562687005
Therefore, approximately, the 90% confidence interval for the students' mean IQ score is between 129.045 - 130.956
Answer:
Step-by-step explanation:
The given expression describes the GP with
<u>The first term:</u>
- t = 80(1.23²⁻²) = 80(1.23⁰) = 80
<u>The common ratio: </u>
It is required to find the sum of the terms from 2 (the bottom number of sigma) through 29 (the top number of sigma)
The number of terms is
The sum of the first 28 terms is
- Sₙ = t(rⁿ - 1)/(r - 1)
- S₂₈ = 80(1.23²⁸ - 1)/(1.23 - 1) = 114125.75573 ≈ 114126 (rounded)
Answer:
P( That it will take over 10 years or more of a year with a rainfall above 50inches) = (0.9938)^10
Step-by-step explanation:
Since the annual rainfall is normally distributed,
Given: that
Mean (µ )= 40
and σ = 4.
Let X be normal random variables of the annual rainfall.
P(that there will be over 10 years or more before a year with a rainfall above 50 inches)
P(>50) = 1-P[X ≤50]
1 - P[X- μ/σ ≤ 50 - 40/4]
=1 - P [Z≤ 5/2]
=1 -Φ(5/2)
=1 - 0.9939
= 0.0062
P( the non occurrence of rainfall above 50 inches)
= 1-0.0062
=0.9938
ASSUMPTION:
P( That it will take over 10 years or more of a year with a rainfall above 50inches) =
