Help solve these using the addition and multiplication properties of equality
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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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