a. Calculate the probability that mean reading age of the sample is greater than 11.65 years

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Let's solve the problems related to reading age. Part (a): Unbiased estimates of mean and variance Given: Sample size, n = 20; Sum of reading ages, ∑ x = 232.6; Sum of squared reading ages, ∑ x² = 2756.22 Step 1: Calculate the unbiased estimate of the mean (μ). The unbiased estimate of the population mean is the sample mean (x). x = (∑ x)/(n); x = 232.6/20; x = 11.63 Rounding to 1 decimal place: x = 11.6 Step 2: Calculate the unbiased estimate of the variance (σ²). The unbiased estimate of the population variance is the sample variance (s²). s² = (∑ x² − ((∑ x)²)/(n))/(n−1); s² = (2756.22 − ((232.6)²)/(20))/(20−1); s² = (2756.22 − 54103.76/20)/(19); s² = (2756.22 − 2705.188)/(19); s² = 51.032/19 s² ≈ 2.68589 Rounding to 1 decimal place: s² = 2.7 Part (b): 95% confidence interval for the mean reading age (μ) Given: Sample size, n = 25; Sample mean, x = 11.4 years; Population variance, σ² = 2.25 (so population standard deviation σ = √(2.25) = 1.5 years); Confidence level = 95% Step 1: Determine the critical z-value. Since the population standard deviation (σ) is known, we use the z-distribution. For a 95% confidence interval, α = 1 − 0.95 = 0.05. We need zα/2 = z0.05/2 = z0.025. From the standard normal distribution table, z0.025 = 1.96. Step 2: Calculate the margin of error (ME). ME = zα/2 × (σ)/(√(n)); ME = 1.96 × (1.5)/(√(25)); ME = 1.96 × 1.5/5; ME = 1.96 × 0.3; ME = 0.588 Step 3: Construct the confidence interval. The confidence interval is given by x ± ME. Lower bound: 11.4 − 0.588 = 10.812; Upper bound: 11.4 + 0.588 = 11.988 The 95% confidence interval for the mean reading age is: (10.812 years, 11.988 years) Part (c): Probability that the sample mean is greater than 11.65 years Given: Population mean, μ = 11.5 years; Population standard deviation, σ = 2.25 years; Sample size, n = 100 We want to find P(X > 11.65). Step 1: Calculate the standard error of the mean (σ_x). σ_x = (σ)/(√(n)); σ_x = (2.25)/(√(100)); σ_x = 2.25/10; σ_x = 0.225 Step 2: Standardize the sample mean (X) to a z-score. Z = X − μσ_x; Z = (11.65 − 11.5)/(0.225); Z = 0.15/0.225 Z ≈ 0.6667 Step 3: Find the probability using the standard normal distribution. We need to find P(Z > 0.6667). Using a z-table or calculator for Z ≈ 0.67: P(Z > 0.67) = 1 − P(Z < 0.67); P(Z > 0.67) = 1 − 0.7486; P(Z > 0.67) = 0.2514 The probability that the mean reading age of the sample is greater than 11.65 years is: 0.2514

A. Calculate the probability that mean reading age of the sample is greater than 11.65 years

Answers

Step-by-step solution

Part a· 2 steps

Calculate the unbiased estimate of the mean ( $\mu$ ).

Calculate the unbiased estimate of the variance ( $\sigma^2$ ).

Part b· 3 steps
$(10.812 \text{ years}, 11.988 \text{ years})$

Determine the critical z-value.

Calculate the margin of error (ME).

Construct the confidence interval.

Part c· 3 steps0.2514

Calculate the standard error of the mean ( $\sigma_{\bar{x}}$ ).

Standardize the sample mean ( $\bar{X}$ ) to a z-score.

Find the probability using the standard normal distribution.

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