Respuesta :
Answer:
Probability that the spending is more than $42.35 is 0.7271.
Step-by-step explanation:
We are given that American Statistical Association budget is distributed normally with a mean spending of $45.67 and a standard deviation of $5.50.
Let X = American Statistical Association budget
So, X ~ N([tex]\mu=45.67,\sigma^{2} =5.5^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{ X -\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean spending = $45.67
[tex]\sigma[/tex] = standard deviation = $5.50
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
So, the probability that the spending is more than $42.35 is given by = P(X > $42.35)
P(X > $42.35) = P( [tex]\frac{ X -\mu}{\sigma}[/tex] > [tex]\frac{42.35-45.67}{5.5}[/tex] ) = P(Z > -0.604) = P(Z < 0.604)
= 0.7271
Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.604 in the z table which will lie between x = 0.60 and x = 0.70 which has an area of 0.7271.
Hence, the probability that the spending is more than $42.35 is 0.7271.
