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'n Opname oor hul voorkeursmanier van oefen, is onder 140 mense in twee ouderdomsgroepe gedoen - NSC Mathematics - Question 11 - 2018 - Paper 1

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'n Opname oor hul voorkeursmanier van oefen, is onder 140 mense in twee ouderdomsgroepe gedoen. Die inligting is hieronder opgesom. OUDERDOM TENNIS HARLOOP GIM TOT... show full transcript

Worked Solution & Example Answer:'n Opname oor hul voorkeursmanier van oefen, is onder 140 mense in twee ouderdomsgroepe gedoen - NSC Mathematics - Question 11 - 2018 - Paper 1

Step 1

Bepaal die waarde van a indien dit gegee word dat 'n tennisvoorkeur en ouderdom onafhanklik van mekaar is.

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Answer

To determine the value of a under the condition that tennis preference and age are independent, we can use the formula for conditional probability:

P(tennisextageextext35)=P(tennis)P(age  ≤ 35)P(age)P(tennis | ext{age} ext{ } ext{ ≤ } 35) = \frac{P(tennis) \cdot P(age \text{ } \text{ ≤ } 35)}{P(age)}

Substituting the values from the table:

21140=a80\frac{21}{140} = \frac{a}{80}

Cross-multiplying gives:

2180=140a21 \cdot 80 = 140 \cdot a

Thus,

a=2180140=12a = \frac{21 \cdot 80}{140} = 12

Therefore, the value of a is 12.

Step 2

Indien dit gegee word dat a = 12, bepaal die waarskynlikheid dat 'n willekeurig gekozen persoon verkiest om gym toe te gaan in die ouderdomsgroep 35 jaar en jonger is.

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Answer

Using the values from the table where a = 12:

We know that:

P(gimextage  ≤ 35)=P(gim)+P(age  ≤ 35)P(gim  and age  ≤ 35)P(gim | ext{age} \text{ }\text{ ≤ } 35) = P(gim) + P(age \text{ }\text{ ≤ } 35) - P(gim \text{ }\text{ and } \text{age} \text{ }\text{ ≤ } 35)

Substituting the values:

=70140+8014040140 = \frac{70}{140} + \frac{80}{140} - \frac{40}{140} =70+8040140=110140 = \frac{70 + 80 - 40}{140} = \frac{110}{140}

This simplifies to:

=1114 or 0.79 = \frac{11}{14} \text{ or } 0.79

Thus, the final result is that the probability is approximately 0.79.

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