The time, in minutes, for Anjan's journey to work on Mondays has mean $$\(38.4\)$$ and standard deviation $$\(6.9\)$$ Find the probability that Anjan's mean journey time for a random sample of 30 Mondays is between 38 and 40 minutes. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_w20_qp_63 Year：2020 Question No：6(a)
Answer:
\(\frac{40-38.4}{\frac{6.9}{\sqrt{30}}}=1.270 \quad \frac{38-38.4}{\frac{6.9}{\sqrt{30}}}=-0.3175\)
\(\Phi\left(\right.\) ' \(\left.1.270^{\prime}\right)-\left(1-\phi\left(' 0.3175^{\prime}\right)\right.\)
\(=0.523(3 \mathrm{sf}) \text { or } 0.522\)
\(\Phi\left(\right.\) ' \(\left.1.270^{\prime}\right)-\left(1-\phi\left(' 0.3175^{\prime}\right)\right.\)
\(=0.523(3 \mathrm{sf}) \text { or } 0.522\)
Knowledge points:
6.4.3 recognise that a sample mean can be regarded as a random variable, and use the facts that
6.4.5 use the Central Limit Theorem where appropriate (Only an informal understanding of the Central Limit Theorem (CLT) is required; for large sample sizes, the distribution of a sample mean is approximately normal.)
Solution:
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