5.6-3 ### Exercises**5.6-1.** Let \( \overline{X} \) be the mean of a random sample of size 12 from the uniform distribution on the interval (0, 1). Approximate \( P(1/2 \leq \overline{X} \leq 2/3) \).**5.6-2.** Let \( Y = X_1 + X_2 + \cdots + X_{15} \) be the sum of a random sample of size 15 from the distribution whose pdf is \( f(x) = \frac{3}{2}x^2, -1 < x < 1 \). Using the pdf of \( Y \), we find that \( P(-0.3 \leq Y \leq 1.5) = 0.22788 \). Use the central limit theorem to approximate this probability.**5.6-3.** Let \( \overline{X} \) be the mean of a random sample of size 36 from an exponential distribution with mean 3. Approximate \( P(2.5 \leq \overline{X} \leq 4) \).**5.6-4.** Approximate \( P(39.75 \leq \overline{X} \leq 41.25) \), where \( \overline{X} \) is the mean of a random sample of size 32 from a distribution with mean \( \mu = 40 \) and variance \( \sigma^2 = 8 \).

The Z-score of a random variable X is defined as follows: Z = (X – µ)/σ. Here, µ and σ are the mean and standard deviation of X, respectively.Consider a random variable X, follows exponential distribution with mean µ = 3. It is known that the mean and standard deviation of exponential distribution are same. That is, σ = 3. According to central limit theorem, for large sample (>30) or if the population follows normal distribution, then the sampling distribution of the sample mean follows normal distribution with the expectation and standard deviation defined as follows: E(x̅) = µ and SD(x̅) = σ/√n. Here, n is the sample size, µ is the population mean and σ is the population standard deviation of the random variable X. Consider that x̅ is the sample mean. Now, for sample of 36, the mean of the sampling distribution of the sample mean is, µx̅ = 3. Now, for sample of 36, the standard deviation of the sampling distribution of the sample mean is, σx̅ = 3/√36 = 0.5. Thus, the sampling distribution of x̅ is approximately normal with mean µx̅ = 3 and standard deviation σx̅ = 0.5. The value of P(2.5 ≤ X̅ ≤ 4) is, Therefore, the value of P(2.5 ≤ X̅ ≤ 4) is 0.8185.