Question

17. A survey of all City Tech students found that 35% say they get their lunch from the salad bar in the cafeteria. A random sample of 140 students is selected and asked if they get their lunch from the salad bar in the cafeteria.

a) What is the probability that at most 55 say they get their lunch from the salad bar in the cafeteria?

b) What is the probability that more than 40 say they get their lunch from the salad bar in the cafeteria?

c) What is the probability that between 50 and 60, inclusive, say they get their lunch from the salad bar in the cafeteria?

Answer #1

P(a person gets lunch from the salad bar in the cafeteria), p = 0.35

q = 1 - p = 0.65

Sample size, n = 140

Mean = np

= 140 x 0.35

= 49

Standard deviation =

=

= 5.6436

P(X < A) = P(Z < (A - mean)/standard deviation)

a) P(at most 55) = P(X < 55.5) (with continuity correction)

= P(Z < (55.5 - 49)/5.6436)

= P(Z < 1.15)

= **0.8749**

b) P(more than 40) = P(X > 40)

= 1 - P(X < 40.5) (with continuity correction)

= 1 - P(Z < (40.5 - 49)/5.6436)

= 1 - P(Z < -1.51)

= 1 - 0.0655

= **0.9345**

c) P(between 50 and 60, inclusive) = P(X < 60.5) - P(X < 49.5)

= P(Z < (60.5 - 49)/5.6436) - P(Z < (49.5 - 49)/5.6436)

= P(Z < 2.04) - P(Z < 0.09)

= 0.9793 - 0.5359

= **0.4434**

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