a company manufactures footballs whose weights are normally distributed with a mean of 410 g and a standard…

a company manufactures footballs whose weights are normally distributed with a mean of 410 g and a standard - deviation of 21 g. (a) what is the probability that a randomly selected football weighs more than 421 g? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage. for parts (b) and (c), even though the sample size is less than 100 we can use the central limit theorem. this is because we are told that the original variable follows a normal distribution and we are given the population standard deviation. (b) what is the probability that a random sample of 30 footballs has a mean weight of more than 421 g? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage. (c) when the company is doing quality - control checks, they take the mean weight of a random sample of 50 footballs. between what two values do the middle 80% of sample mean weights of 50 footballs lie? round your answers to 2 decimal places. from g to g

a company manufactures footballs whose weights are normally distributed with a mean of 410 g and a standard - deviation of 21 g. (a) what is the probability that a randomly selected football weighs more than 421 g? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage. for parts (b) and (c), even though the sample size is less than 100 we can use the central limit theorem. this is because we are told that the original variable follows a normal distribution and we are given the population standard deviation. (b) what is the probability that a random sample of 30 footballs has a mean weight of more than 421 g? round your answer to 4 decimal places. make sure that your answer is a probability, not a percentage. (c) when the company is doing quality - control checks, they take the mean weight of a random sample of 50 footballs. between what two values do the middle 80% of sample mean weights of 50 footballs lie? round your answers to 2 decimal places. from g to g

Answer

Explanation:

Step1: Calculate z - score for part (a)

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 421$, $\mu=410$, and $\sigma = 21$. $z=\frac{421 - 410}{21}=\frac{11}{21}\approx0.5238$ We want $P(X>421)$, which is $1 - P(X\leq421)$. Looking up the z - score in the standard normal table, $P(Z\leq0.5238)\approx0.6990$. So $P(X > 421)=1 - 0.6990 = 0.3010$.

Step2: Calculate z - score for part (b)

For the sampling distribution of the sample mean, the standard deviation of the sample mean (also known as the standard error) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 21$, $n = 30$. So $\sigma_{\bar{x}}=\frac{21}{\sqrt{30}}\approx3.834$. The z - score for $\bar{x}=421$ is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}=\frac{421 - 410}{3.834}\approx2.87$. We want $P(\bar{X}>421)=1 - P(\bar{X}\leq421)$. Looking up the z - score in the standard normal table, $P(Z\leq2.87)\approx0.9979$. So $P(\bar{X}>421)=1 - 0.9979 = 0.0021$.

Step3: Find z - scores for part (c)

For the middle 80% of the distribution, the tails on both sides have a combined area of $1 - 0.80=0.20$, so the area in each tail is $\frac{0.20}{2}=0.10$. The z - score corresponding to the left - hand tail area of 0.10 is $z_{1}\approx - 1.28$ and the z - score corresponding to the right - hand tail area of 0.10 is $z_{2}\approx1.28$. The standard deviation of the sample mean for $n = 50$ is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{21}{\sqrt{50}}\approx2.97$. The lower value $L=\mu+z_{1}\sigma_{\bar{x}}=410+( - 1.28)\times2.97=410-3.80 = 406.20$. The upper value $U=\mu+z_{2}\sigma_{\bar{x}}=410 + 1.28\times2.97=410+3.80 = 413.80$.

Answer:

(a) 0.3010 (b) 0.0021 (c) 406.20, 413.80