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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

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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 cou... show full transcript

Worked Solution & Example Answer: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

Step 1

Find the value of k.

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Answer

To find the value of k, we start by using the standard normal distribution. We know that 1% of the females are shorter than k, which corresponds to a z-score of -2.3263 (from z-score tables). We can standardize k using the formula:

P(Z<z)=P(X−μσ<z)P(Z < z) = P\left( \frac{X - \mu}{\sigma} < z \right)

Where:

  • ( X ) is the height,
  • ( \mu = 166.5 ) cm is the mean,
  • ( \sigma = 6.1 ) cm is the standard deviation.

Setting this up, we have:

−2.3263=k−166.56.1-2.3263 = \frac{k - 166.5}{6.1}

To find k, we rearrange the equation:

k=166.5+(−2.3263×6.1)k = 166.5 + (-2.3263 \times 6.1)

Calculating this gives:

k≈152.309extcmext(roundedto152or152.3)k \approx 152.309 ext{ cm} ext{ (rounded to 152 or 152.3)}

Step 2

Find the proportion of females from this country with heights between 150 cm and 175 cm.

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Answer

To find the proportion of females with heights between 150 cm and 175 cm, we need to convert these heights into z-scores.

For 150 cm:

Z150=150−166.56.1≈−2.677Z_{150} = \frac{150 - 166.5}{6.1} \approx -2.677

For 175 cm:

Z175=175−166.56.1≈1.402Z_{175} = \frac{175 - 166.5}{6.1} \approx 1.402

Now, we use the z-tables to find:

  • P(Z < -2.677) and P(Z < 1.402).

Calculating these values:

P(150<X<175)=P(Z<1.402)−P(Z<−2.677)≈0.9148−0.0037=0.9111ext(approximately0.915)P(150 < X < 175) = P(Z < 1.402) - P(Z < -2.677) \approx 0.9148 - 0.0037 = 0.9111 ext{ (approximately 0.915)}

Step 3

Find the probability that her height is more than 160 cm.

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Answer

We are interested in finding the probability that a randomly selected female's height is more than 160 cm from the range of 150 cm to 175 cm.

Calculating the z-score for 160 cm:

Z160=160−166.56.1≈−1.061Z_{160} = \frac{160 - 166.5}{6.1} \approx -1.061

Now we calculate the probability:

P(X>160∣150<X<175)=1−P(Z<−1.061)≈1−0.1446=0.8554ext(approximately0.847)P(X > 160 | 150 < X < 175) = 1 - P(Z < -1.061) \approx 1 - 0.1446 = 0.8554 ext{ (approximately 0.847)}

Step 4

Carry out a suitable test to assess Mia’s belief.

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Answer

We will carry out a hypothesis test.

  • Null Hypothesis (H₀): ( \mu = 166.5 ) cm (the mean height is equal to 166.5 cm)
  • Alternative Hypothesis (H₁): ( \mu < 166.5 ) cm (the mean height is less than 166.5 cm)

We will use a significance level of 5%. Given that the sample mean ( \bar{x} ) is 164.6 cm, and a sample size of n = 50:

We calculate the standard error (SE):

SE=σn=7.450≈1.046SE = \frac{\sigma}{\sqrt{n}} = \frac{7.4}{\sqrt{50}} \approx 1.046

Next, we calculate the z-score:

Z=xˉ−μSE=164.6−166.51.046≈−1.818Z = \frac{\bar{x} - \mu}{SE} = \frac{164.6 - 166.5}{1.046} \approx -1.818

Now we refer to the z-table for a z-score of -1.818:

  • The p-value obtained is approximately 0.0347.

Since 0.0347 < 0.05, we reject the null hypothesis. There is evidence to support Mia’s belief that the mean height is less than 166.5 cm.

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