e to nomal? Should P Why or why not? tribution has a m a. What is b. Estimate (u

Question

e to nomal? Should P Why or why not? tribution has a m a. What is b. Estimate (using area) the relative freque the total area under the curve! of Figure s.15 e. Estimate the relative frequency of values grear d. Estimate the relative frequency of values les e. Estimate the relative frequency of values ber less than 10. o alues greater than 1 than 11 between 1 PROJ 72 27 SA bi and 110 1.9 14S 160 40 55 70 85 100 15 130 m for the The mean e to nor Figure 5.13 Q scores n? Why 25. Estimating Areas. Consider the graph of the normal dis e freqencis n tion in Figure 5.14, which illustrates the relative freq a distribution of systolic blood pressures for a sample of s s in students. The distribution has a standard deviation of 14 ale a. What is the mean of the distribution? b. Estimate (using area) the percentage of students wh blood pressure is less than 100. ‘ c. Estimate the percentage of students whose bl pressure is between 110 and 130. 0od d. Estimate the percentage of students whose blon pressure is greater than 130. 55 7085 100 115 130 145 160 175 Blood pressure gure 5.14

Explanation / Answer

Question 24

(a) Total area under curve = 1 [Propoerty of probability distribution ]

(b) Mean = 100 ; Standard deviation = 16

Relative frequency for less than 100 = 0.50

(c) Relative frequency for values greater than 110

Pr(X > 110) = Pr(X > 110 ; 100 ; 16) = 1 - Pr(X < 110 ; 100; 16)

Z = (110 - 100)/16 = 0.625

Pr(X > 110) = 1 - Pr(Z < 0.625) = 1 - 0.7340 = 0.2660

(d) Pr(X < 110) = Pr(X < 110 ; 100 ; 16) = 0.7340

(e) Pr(100 < X < 110) = Pr(X <110) - Pr(X <100) = 0.7340 = 0.5 = 0.2340

Queston2 5

(a) Mean of the distribution = 114 (As we estimate it from the graph)

(b) Pr(X < 100) = ? where X is the systolic blood pressure of a girl student.

Pr(X < 100) = Pr(X < 100 ; 114; 14)

Z = (100 - 114)/14 = -1

Pr(X < 100) = Pr(Z < -1) = 0.1587

(c) Pr(100 < X < 130) = Pr(X < 130) - Pr(X < 100)

Z2 = (130 - 114)/14 = 1.1429 ; Z1 = -1

Pr(100 < X < 130) = Pr(X < 130) - Pr(X < 100) = Pr(Z2 < 1.1429) - Pr(Z1 < -1)

= 0.8735 - 0.1587

= 0.7148

(d) Pr(X > 130) = 1 - Pr(X < 130) = 1 - 0.8735 = 0.1265

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