A particular piece of music was played by 91 pianists and for each pianist, the number of incorrect notes was recorded. The results are summarised in the table. Calculate an estimate for the mean number of incorrect notes. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_w20_qp_53 Year：2020 Question No：7(c)
Answer:
Midpoints: \( 3\quad8\quad 15.5\quad30 .5\quad55 .5 \)
\( \begin{array}{l}
\text { Mean }=\frac{3 \times 10+8 \times 5+15.5 \times 26+30.5 \times 32+55.5 \times 18}{91} \\=\frac{30+40+403+976+999}{91} \\=\frac{2448}{91} \\26.9, \quad 26 \frac{82}{91}
\end{array}
\)
\( \begin{array}{l}
\text { Mean }=\frac{3 \times 10+8 \times 5+15.5 \times 26+30.5 \times 32+55.5 \times 18}{91} \\=\frac{30+40+403+976+999}{91} \\=\frac{2448}{91} \\26.9, \quad 26 \frac{82}{91}
\end{array}
\)
Knowledge points:
5.1.5 calculate and use the mean and standard deviation of a set of data (including grouped data) either from the data itself or from given totals , and use such totals in solving problems which may involve up to two data sets.
Solution:
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