Q4. (6 pts) A grocery owner purchased 20000 pears and sampled 50 of them (withou

Question

Q4. (6 pts) A grocery owner purchased 20000 pears and sampled 50 of them (without replacement); the average of these samples is 112 grams and the standard deviation of these samples is 4 grams. What is the probability to have these sampled pears between 106 and 118 grams? And also give the number of pears. If the average of 50 pears is equal to that of population and the standard deviation of whole population could also be transferred) what is the probability to have pears between 100 and 124 grams (whole population)? And how many pears of whole population have weight between 100 and 124 grams? 1. 2. 3.

Explanation / Answer

Here average of the sample = 112 grams

standard deviation of these samles = 4 grams

(a) Here if a random pear has x gram of weight.

Pr(106 gm < x < 118 gms) = Pr( x < 118 gms) - Pr(x < 106 gms)

Z2 = (118 - 112)/ 4 = 1.5 ; Z1 = (106 - 112)/4 = -1.5

Pr(106 gm < x < 118 gms) = Pr( x < 118 gms) - Pr(x < 106 gms) = Pr(Z < 1.5) - Pr(Z < -1.5)

= 0.9332 - 0.0668

= 0.8664

Number of such pears = 50 * 0.8664 = 43.32 or say 43 /44

(2) Average weight of population = average sample weight = 112 gms

standard deviation of population = 4 gms

Pr(100 gms < x < 124 gms) = Pr( x < 124 gms) - Pr(x < 100 gms)

Z2 = (124 - 112)/ 4 = 3 ; Z1 = (100- 112)/4 = -3

= Pr(Z < 3) - Pr(Z < -3)

= 0.99865 - 0.00135 = 0.9973

(3) Number of peers hae weight between 100 and 124 grams = 0.9973 * 20000 = 19946

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.