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
13 divided by 11.3 is 1.150442477876106. So 15 by 1.150442477876106 and you get 13.03846162772485. Just round it to the nearest tenth and you get 13.0.
Answer:
I need this answer too !!!
Answer:d
Step-by-step explanation:
Step-by-step explanation:
You have 5 digits.Youdonot say if duplication is allowed. I will assume not
For the first digit you have 5 choices. One of the 5 digits is gone.
You now have 4 digits to choose from. You pick one. Now you have but 3 left.
The total answer if 5*4*3*2*1 = 120
