Question

Chapter 7 Supplemental Practice Problems Mallu 6:7 " 2 number of standard deviations between x (variable of interest) and mu f 114 ounces and a The weight of a McDonald's cheesebureer is normally distributed with a mean of 114 ounces and a standard deviation of 7 ounces. Using the z table compute the following probabilities. a) P/x<113.25 ounces) or be i den b) P(x>113.25 ounces) c) P(112.75<x<113.85) a gost d) P(x<115.75 ounces) dolgoz nihy W o e) P(113.25<x<115.75 ounces) f) P(x>115.75) R s nio 8) P(115.75<x<116.85) Using the T183-84 to compute probabilities: Keep in mind all CDFs use the value and all the area below a given point. To solve for a) above do the following: Select the 2nd key and select the vars key to activate distributions. Select normalcdf for lower leave the lower value set to -1E99 for the upper: enter 113.25, enter the mean and the standard deviation and select enter until the probability of .457 shows in the calculator window.

Answer 1

α. Ρ(x < 113.25) = Φ(113.5 – 114)/7) = Φ(-0.07) = 1 - Φ(0.07) = 0.4721 6. Ρ(r> 113.25) = 1 – P(x < 113.25) = 1 – 0.4721 = 0.5279 C. Ρ(112.75 < < 113.85) = Φ(113.85 – 114)/7) - Φ((112.75 – 114)/7) = Φ(-0.02) – Φ(-0.18) = Φ(0.18) – Φ(0.02) = 0.5714 - 0.5080 = 0.0634 d. P(x < 115.75) = Φ((115.75 – 114)/7) = Φ(0.25) = 0.5987 €. Ρ(113.25 <1 <115.75) = P(x < 115.75) – Pr < 113.25) = 0.5987 – 0.4721 = 0.1266 f. Ρ(.>115.75) = 1 – P(< 115.75) = 1 – 0.5987 = 0.4013 9. Ρ(115.75 < < 116.85) = Φ((116.85 – 114)/7) - P(x <115.75) = Φ(0.41) - 0.5987 = 0.6591 – 0.5987 = 0.0604

where (2) = cdf of standard normal distribution, 0(-)=1-02) and since z is continuous random variable so P(Z < 2) = P(Z < 2) = 0(2).