A simple random sample of size n=36 is obtained from a population that is skewed right with mu=80 and sigma=24. (a) Describe the sampling distribution of bar(x). (b) What is P( bar(x) > 86.8) ? (c) What is P( bar(x) <= 70.2) ? (d) What is P(74 < bar(x) < 89.6) ?

Identify Distribution: Identify the distribution of the sample mean (bar(x)) for a large sample size from a skewed population. Since n=36 is sufficiently large, by the Central Limit Theorem, bar(x) will be approximately normally distributed with mean mu and standard deviation sigma/sqrt(n). mu = 80, sigma = 24, n = 36. Standard deviation of bar(x) = 24 / sqrt(36) = 4.Calculate P(>86.8): Calculate P(bar(x) > 86.8). First, convert 86.8 to a z-score. z = (86.8 - 80) / 4 = 1.7. Using the z-table, P(Z > 1.7) ≈ 0.0446.Calculate P(<=70.2): Calculate P(bar(x) <= 70.2). Convert 70.2 to a z-score. z = (70.2 - 80) / 4 = -2.45. Using the z-table, P(Z <= -2.45) ≈ 0.0071.Calculate P(74 Calculate P(74 < bar(x) < 89.6). Convert 74 and 89.6 to z-scores. z1 = (74 - 80) / 4 = -1.5, z2 = (89.6 - 80) / 4 = 2.4. Using the z-table, P(-1.5 < Z < 2.4) ≈ P(Z < 2.4) - P(Z < -1.5) ≈ 0.9918 - 0.0668 = 0.925.

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A simple random sample of size
n=36 is obtained from a population that is skewed right with
mu=80 and
sigma=24.
(a) Describe the sampling distribution of
bar(x).
(b) What is
P( bar(x) > 86.8) ?
(c) What is
P( bar(x) <= 70.2) ?
(d) What is
P(74 < bar(x) < 89.6) ?

A simple random sample of size $n=36$ is obtained from a population that is skewed right with $\mu=80$ and $\sigma=24$ . $\newline$ (a) Describe the sampling distribution of $\bar{x}$ . $\newline$ (b) What is P(\bar{x} > 86.8)? $\newline$ (c) What is $P(\bar{x} \leq 70.2)$ ? $\newline$ (d) What is P(74 < \bar{x} < 89.6)?

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Q. A simple random sample of size

n=36

is obtained from a population that is skewed right with

\mu=80

and

\sigma=24

\newline

(a) Describe the sampling distribution of

\bar{x}

\newline

(b) What is

P(\bar{x} > 86.8)

\newline

P(\bar{x} \leq 70.2)

\newline

(d) What is

P(74 < \bar{x} < 89.6)

Identify Distribution: Identify the distribution of the sample mean $\bar{x}$ for a large sample size from a skewed population. Since $n=36$ is sufficiently large, by the Central Limit Theorem, $\bar{x}$ will be approximately normally distributed with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$ . $\mu = 80$ , $\sigma = 24$ , $n = 36$ . Standard deviation of $\bar{x}$ = $24 / \sqrt{36}$ = $n=36$ $0$ .
Calculate P(>86.8): Calculate P(\bar{x} > 86.8). $\newline$ First, convert $86.8$ to a z-score. $\newline$ $z = \frac{86.8 - 80}{4} = 1.7$ . $\newline$ Using the z-table, P(Z > 1.7) \approx 0.0446.
Calculate $P(\leq 70.2)$ : Calculate $P(\bar{x} \leq 70.2)$ . Convert $70.2$ to a z-score. $z = \frac{70.2 - 80}{4} = -2.45$ . Using the z-table, $P(Z \leq -2.45) \approx 0.0071$ .
Calculate P(74<\bar{x}<89.6): Calculate P(74 < \bar{x} < 89.6). Convert $74$ and $89.6$ to z-scores. $z_1 = \frac{74 - 80}{4} = -1.5$ , $z_2 = \frac{89.6 - 80}{4} = 2.4$ . Using the z-table, P(-1.5 < Z < 2.4) \approx P(Z < 2.4) - P(Z < -1.5) \approx 0.9918 - 0.0668 = 0.925.