All 303 families

RV-17-001 · Frequency table with an unknown count

The table shows the number of repairs needed by each bicycle.
Repairs012
Frequency9n13
The mean number of repairs is 2119\frac{21}{19}.

(a) Find n.

(b) One bicycle is selected at random. Find the probability that it needed at least one repair.

Part (a) Mean from a Frequency Table with Algebraic Unknown Booklet p. 11 & p. 67
Method
1) Use the actual value x for ungrouped data; use a midpoint only for a class interval
2) Multiply each value by its frequency
3) Divide the total fx (or fm) by the total frequency
Worked Example (RV-17-001 · Part (a))
Find n.
1)
Repairs xFrequency ffx
090
1nn
21326
Totaln+22n+22n+26n+26
2)
n+26n+22=2119\frac{n+26}{n+22}=\frac{21}{19}
3)
Multiply both sides by the denominators: 19(n+26)=21(n+22)19(n+26)=21(n+22).
4)
19n+494=21n+462⇒−2n=−32⇒n=1619n+494=21n+462\Rightarrow -2n=-32\Rightarrow n=16
Ans = 16
Part (b) Mean from a Frequency Table with Algebraic Unknown Booklet p. 60
Method
1) Count the bicycles with one or two repairs
2) Divide by the total number of bicycles
Worked Example (RV-17-001 · Part (b))
One bicycle is selected at random. Find the probability that it needed at least one repair.
1)
From part (a), n=16n=16.
2)
P(at least one repair)=16+139+16+13=2938P(\text{at least one repair})=\frac{16+13}{9+16+13}=\frac{29}{38}
Ans = 2938\frac{29}{38}