Answer: 0.0793
Step-by-step explanation:
Let the IQ of the educated adults be X then;
Assume X follows a normal distribution with mean 118 and standard deviation of 20.
This is a sampling question with sample size, n =200
To find the probability that the sample mean IQ is greater than 120:
P(X > 120) = 1 - P(X < 120)
Standardize the mean IQ using the sampling formula : Z = (X - μ) / σ/sqrt n
Where; X = sample mean IQ; μ =population mean IQ; σ = population standard deviation and n = sample size
Therefore, P(X>120) = 1 - P(Z < (120 - 118)/20/sqrt 200)
= 1 - P(Z< 1.41)
The P(Z<1.41) can then be obtained from the Z tables and the value is 0.9207
Thus; P(X< 120) = 1 - 0.9207
= 0.0793
Answer:
m = - 1
Step-by-step explanation:
9m + 13 = 4
subtract 13 from both sides
9m = 4 - 13
9m = - 9
divide both sides by 9
m = - 1 (negative 1)
Answer:
s = 21.16
Step-by-step explanation:
0.5s +1= 7+ 4.58
Combine like terms
0.5s +1=11.58
Subtract 1 from each side
0.5s +1-1= 11.58-1
.5s = 10.58
Multiply by 2
.5s *2 = 10.58*2
s = 21.16
Answer:
ita A hope it helps
Step-by-step explanation:
