Question

American Statistical Association budget is distributed normally with a mean spending of $45.67 and a standard deviation of $5.50. What is the probability that the spending is more than $42.35

Answer 1

Probability that the spending is more than $42.35 is 0.7271.

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(
`\mu=45.67,\sigma^(2) =5.5^(2)`)

The z-score probability distribution for normal distribution is given by;

Z =
`( X -\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = mean spending = $45.67

`\sigma` = 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(
`( X -\mu)/(\sigma)` >
`(42.35-45.67)/(5.5)` ) = P(Z > -0.604) = P(Z < 0.604)

= 0.7271

Now, in the z table the P(Z
`\leq` 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.