Answer:
a) 0.2318
b) 0.2609
c) No it is not unusual for a broiler to weigh more than 1610 grams
Step-by-step explanation:
We solve using z score formula
z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.
(a) What proportion of broilers weigh between 1150 and 1308 grams?
For 1150 grams
z = 1150 - 1387/192
= -1.23438
Probability value from Z-Table:
P(x = 1150) = 0.10853
For 1308 grams
z = 1308 - 1387/192
= -0.41146
Probability value from Z-Table:
P(x = 1308) = 0.34037
Proportion of broilers weigh between 1150 and 1308 grams is:
P(x = 1308) - P(x = 1150)
0.34037 - 0.10853
= 0.23184
≈ 0.2318
(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?
1510 - 1387/192
= 0.64063
Probabilty value from Z-Table:
P(x<1510) = 0.73912
P(x>1510) = 1 - P(x<1510) = 0.26088
≈ 0.2609
(c) Is it unusual for a broiler to weigh more than 1610 grams?
1610- 1387/192
= 1.16146
Probability value from Z-Table:
P(x<1610) = 0.87727
P(x>1610) = 1 - P(x<1610) = 0.12273
≈ 0.1227
No it is not unusual for a broiler to weigh more than 1610 grams
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