The heights of females from a country are normally distributed with - a mean of 166.5 cm - a standard deviation of 6.1 cm Given that 1% of females from this country are shorter than $k$, (a) Find the value of $k$ - Edexcel - A-Level Maths Statistics - Question 5 - 2021 - Paper 1

To find the proportion of females with heights between 150 cm and 175 cm, we can find:

$P(150 < X < 175) = P(X < 175) - P(X < 150)$

Converting to Z-scores:

$P(X < 175) = P\left(Z < \frac{175 - 166.5}{6.1}\right) = P(Z < 1.4033) \approx 0.9194$

$P(X < 150) = P\left(Z < \frac{150 - 166.5}{6.1}\right) = P(Z < -2.6772) \approx 0.0037$

Thus,

$P(150 < X < 175) \approx 0.9194 - 0.0037 \approx 0.9157$

Therefore, the proportion is approximately 0.915.

We need to find:

$P(X > 160 | 150 < X < 175)$

Using conditional probability,

$P(X > 160 | 150 < X < 175) = \frac{P(160 < X < 175)}{P(150 < X < 175)}$

Calculate $P(160 < X < 175)$:

$P(160 < X < 175) = P(X < 175) - P(X < 160)$

Calculate the Z-score for 160:

$P(X < 160) = P\left(Z < \frac{160 - 166.5}{6.1}\right) \approx P(Z < -1.0349) \approx 0.1497$

So,

$P(160 < X < 175) \approx 0.9194 - 0.1497 \approx 0.7697$

Thus,

$P(X > 160 | 150 < X < 175) \approx \frac{0.7697}{0.9157} \approx 0.8407$

We need to conduct a hypothesis test:

• Null Hypothesis ($H_0$): $\mu = 166.5$ cm
• Alternative Hypothesis ($H_1$): $\mu < 166.5$ cm

Using the sample mean ($\bar{x} = 164.6$ cm) and a sample size of $n = 50$, we calculate the test statistic:

$Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{164.6 - 166.5}{7.4 / \sqrt{50}} \approx -2.035$

Now, we compare the test statistic to the critical value:

For a significance level of 0.05, the critical value from the Z-table is approximately -1.645.

Since -2.035 < -1.645, we reject the null hypothesis.

There is evidence to support Mia’s belief that the mean height of females from this country is less than 166.5 cm.