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  

Step-by-step 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 = $\frac{X-\mu}{\sigma}$

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

b. P(X<100) = P( Z < $\frac{100-527}{160}$) = 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<= $\frac{450-527}{160}$) = 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<= $\frac{300-527}{160}$) = 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.