Y=5x² – 1 Please answer I need asap thank you!!
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Answer:
a) Standard error = 2
b) Range = (76.08, 83.92)
c) P=0.69
d) Smaller
e) Greater
Step-by-step explanation:
a) When we have a sample taken out of the population, the standard error of the mean is calculated as:
where n is te sample size (n=25) and σ is the population standard deviation (σ=10).
Then, the standard error of the classroom average score is 2.
b) The calculations for this range are the same that for the confidence interval, with the difference that we know the population mean.
The population standard deviation is know and is σ=10.
The population mean is M=80.
The sample size is N=25.
The standard error of the mean is σM=2.
The z-value for a 95% confidence interval is z=1.96.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:
The range that we expect the average classroom test score to fall 95% of the time is (76.08, 83.92).
c) We can calculate this by calculating the z-score of X=79.
Then, the probability of getting a average score of 79 or higher is:
The approximate probability that a classroom will have an average test score of 79 or higher is 0.69.
d) If the sample is smaller, the standard error is bigger (as the square root of the sample size is in the denominator), so the spread of the probability distribution is more. This results then in a smaller probability for any range.
e) If the population standard deviation is smaller, the standard error for the sample (the classroom) become smaller too. This means that the values are more concentrated around the mean (less spread). This results in a higher probability for every range that include the mean.
The probability that the SRS of 10 students will spend an average of between 600 and 700 dollars is 0.8132.
Let x be the total amount spent by students.
x follows normal distribution with mean μ = 650,
standard deviation δ = 120
We take a simple random sample of size n = 10
We are asked to find average spending of 10 students ( ) is between 600 and 700
We have to find P( 600 <= <= 700)
According to the sampling distribution of the sample mean , it follows an approximately normal distribution with mean μ{
} = μ and standard deviation δ{
} = {δ}/{√n}
Therefore here mean of ,( μ{
} ) = 650 and
standard deviation of , δ{
} = {120}/{√10} = 37.9473
The probability that the SRS of 10 students will spend an average of between 600 and 700 dollars is,
Let Z= x - μ / δ
Z₁ = 600 - 650 / 37.9473 = -1.32 similarly
Z₂ = 700 - 650 / 37.9473 = 1.32
From standard normal distribution table, P( -1.32 < < 1.32) = 0.8132
The probability that the SRS of 10 students will spend an average of between 600 and 700 dollars is 0.8132
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