student scores on a standardized test are normally distributed with mean 540 and standard deviation…
student scores on a standardized test are normally distributed with mean 540 and standard deviation 120.\n(a) what is the probability that a randomly selected student scores below 750 on the test? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage.\n(b) what is the probability that a randomly selected student scores above 666 on the test? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage.\n(c) what is the probability that a randomly selected student scores between 294 and 450 on the test? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage.\n(d) what is the 70th percentile of scores on the exam? round your answer to two decimal places.\n(e) what is the probability that a randomly sample of 131 students has an average score below 538? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage.\n(f) what is the probability that a randomly sample of 131 students has an average score above 548? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage.\n(g) what is the probability that a randomly sample of 131 students has an average score between 529 and 552? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage.
Answer
Explanation:
Step1: Standardize the value
For part (a), we use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $\mu = 540$, $\sigma=120$, and $x = 750$. So $z=\frac{750 - 540}{120}=\frac{210}{120}=1.75$. Then we find $P(X<750)=P(Z < 1.75)$. Looking up in the standard - normal table, $P(Z < 1.75)=0.9599$.
Step2: Standardize for part (b)
For $x = 666$, $z=\frac{666 - 540}{120}=\frac{126}{120}=1.05$. Then $P(X>666)=1 - P(X\leqslant666)=1 - P(Z\leqslant1.05)$. From the standard - normal table, $P(Z\leqslant1.05)=0.8531$, so $P(X>666)=1 - 0.8531 = 0.1469$.
Step3: Standardize for part (c)
For $x_1 = 294$, $z_1=\frac{294 - 540}{120}=\frac{- 246}{120}=-2.05$. For $x_2 = 450$, $z_2=\frac{450 - 540}{120}=\frac{-90}{120}=-0.75$. Then $P(294<X<450)=P(-2.05<Z<-0.75)=P(Z < - 0.75)-P(Z < - 2.05)$. From the standard - normal table, $P(Z < - 0.75)=0.2266$ and $P(Z < - 2.05)=0.0202$, so $P(294<X<450)=0.2266 - 0.0202=0.2064$.
Step4: Find the z - score for the 70th percentile
For part (d), we need to find the z - score $z$ such that $P(Z < z)=0.70$. Looking up in the standard - normal table, the closest $z$ value is approximately $z = 0.52$. Then we use the formula $z=\frac{x-\mu}{\sigma}$ to solve for $x$. So $x=\mu+z\sigma=540+0.52\times120=540 + 62.4=602.40$.
Step5: Use the central limit theorem for part (e)
The standard deviation of the sample mean $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $n = 131$ and $\sigma = 120$. So $\sigma_{\bar{x}}=\frac{120}{\sqrt{131}}\approx10.47$. For $\bar{x}=538$, $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}=\frac{538 - 540}{10.47}\approx - 0.19$. Then $P(\bar{X}<538)=P(Z < - 0.19)=0.4247$.
Step6: Use the central limit theorem for part (f)
For $\bar{x}=548$, $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}=\frac{548 - 540}{10.47}\approx0.76$. Then $P(\bar{X}>548)=1 - P(\bar{X}\leqslant548)=1 - P(Z\leqslant0.76)$. From the standard - normal table, $P(Z\leqslant0.76)=0.7764$, so $P(\bar{X}>548)=1 - 0.7764 = 0.2236$.
Step7: Use the central limit theorem for part (g)
For $\bar{x}_1 = 529$, $z_1=\frac{\bar{x}1-\mu}{\sigma{\bar{x}}}=\frac{529 - 540}{10.47}\approx - 1.05$. For $\bar{x}_2 = 552$, $z_2=\frac{\bar{x}2-\mu}{\sigma{\bar{x}}}=\frac{552 - 540}{10.47}\approx1.15$. Then $P(529<\bar{X}<552)=P(-1.05<Z<1.15)=P(Z < 1.15)-P(Z < - 1.05)$. From the standard - normal table, $P(Z < 1.15)=0.8749$ and $P(Z < - 1.05)=0.1469$, so $P(529<\bar{X}<552)=0.8749 - 0.1469=0.7280$.
Answer:
(a) 0.9599 (b) 0.1469 (c) 0.2064 (d) 602.40 (e) 0.4247 (f) 0.2236 (g) 0.7280