Question

The average amount parents and children spent per child on back-to-school clothes in Autumn 2010 was $527. Assume the standard deviation is $160 and that the amount spent is normally distributed. a. What is the probability that the amount spent on a randomly selected child is more than $700?b. What is the probability that the amount spent on a randomly selected child is less than $100?c. What is the probability that the amount spent on a randomly selected child is between $450 and $700? (Round to four decimal places)d. What is the probability that the amount spent on a randomly selected child is no more than $300? (Round to four decimal places)

Answer 1

Answer: a. 0.14085

b. 3.826 x
`10^(-3)`

c. 0.5437

d. 0.0811

Explanation:

Given average amount parents and children spent per child on back-to-school clothes in Autumn 2010 ,
`\mu` = $527

Given standard deviation ,
`\sigma` = $160

Let X = amount spent on a randomly selected child

Also Z =
`(X-\mu)/(\sigma)`

a. Probability(X>$700) = P(
`(X-\mu)/(\sigma)` >
`(700-527)/(160)`) = P(Z>1.08125) = 0.14085 {Using Z % table}

b. P(X<100) = P( Z <
`(100-527)/(160)`) = P(Z< -2.66875) = P(Z > 2.66875) = 3.826 x
`10^(-3)`

c. P(450<X<700) = P(X<700) - P(X<=450)

P(X<700) = 1 - P(X>=700) = 1 - 0.14085 = 0.8592

P(X<=450) = P(Z<=
`(450-527)/(160)`) = P(Z<= -0.48125) = P(Z<=0.48125) = 0.3155

So final P(450<X<700) = 0.8592 - 0.3155 = 0.5437

d. P(X<=300) = P(Z<=
`(300-527)/(160)`) = P(Z<= -1.4188) = P(Z>=1.4188) = 0.0811

All the above probabilities are calculated using Z % table along with interpolation between two values.