Question

the heights of a group of people are normally distributed with a mean of 174 cm and a standard deviation of 6 cm. find the probability that a person selected at random (a) is at least 170 cm tall (b) is no taller than 180 cm a) z - score= b) z - score= percentage/probability = percentage/probability = batteries for a radio have a mean life of 160 hours and a standard deviation of 30 hours. assuming the battery life follows a normal distribution, calculate: (a) the proportion of batteries which have a life between 150 and 180 hours z - score= percentage/probability = the weights of bags of red gravel may be modelled by a normal distribution with mean 25.8kg and standard deviation 0.5kg. (a) determine the probability that a randomly selected bag of red gravel will weigh (i) less than 25kg. (ii) between 25.3kg and 26.5kg (b) determine, to two decimal places, the weight exceeded by 75% of bags i) z - score= ii) z - score= percentage/probability = percentage/probability =

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean and $\sigma$ is the standard deviation.

Step2: Solve part (a) of the first problem (heights)

Given $\mu = 174$ cm, $\sigma=6$ cm and $x = 170$ cm.
$z=\frac{170 - 174}{6}=\frac{- 4}{6}\approx - 0.67$
We want $P(X\geq170)=1 - P(X < 170)$. Looking up $z=-0.67$ in the standard - normal table, $P(Z < - 0.67)=0.2514$. So $P(X\geq170)=1 - 0.2514 = 0.7486$.

Step3: Solve part (b) of the first problem (heights)

Given $x = 180$ cm, $\mu = 174$ cm and $\sigma = 6$ cm.
$z=\frac{180 - 174}{6}=1$
Looking up $z = 1$ in the standard - normal table, $P(Z<1)=0.8413$.

Step4: Solve part (a) of the second problem (batteries)

Given $\mu = 160$ hours, $\sigma = 30$ hours.
For $x_1 = 150$ hours, $z_1=\frac{150 - 160}{30}=\frac{-10}{30}\approx - 0.33$.
For $x_2 = 180$ hours, $z_2=\frac{180 - 160}{30}=\frac{20}{30}\approx0.67$.
$P(-0.33

Step5: Solve part (a)(i) of the third problem (gravel)

Given $\mu = 25.8$ kg, $\sigma = 0.5$ kg and $x = 25$ kg.
$z=\frac{25 - 25.8}{0.5}=\frac{-0.8}{0.5}=-1.6$
Looking up $z=-1.6$ in the standard - normal table, $P(Z < - 1.6)=0.0548$.

Step6: Solve part (a)(ii) of the third problem (gravel)

For $x_1 = 25.3$ kg, $z_1=\frac{25.3 - 25.8}{0.5}=\frac{-0.5}{0.5}=-1$.
For $x_2 = 26.5$ kg, $z_2=\frac{26.5 - 25.8}{0.5}=\frac{0.7}{0.5}=1.4$.
$P(-1

Step7: Solve part (b) of the third problem (gravel)

If $75\%$ of the bags exceed a certain weight, then the area to the left of this weight is $1 - 0.75=0.25$.
Looking up the $z$ - value corresponding to an area of $0.25$ in the standard - normal table, $z\approx - 0.67$.
Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we can solve for $x$:
$-0.67=\frac{x - 25.8}{0.5}$
$x-25.8=-0.67\times0.5=-0.335$
$x=25.8-0.335 = 25.465\approx25.47$ kg.

Answer:

• First problem (heights):
• (a) Z - Score: $-0.67$, Percentage/Probability: $0.7486$
• (b) Z - Score: $1$, Percentage/Probability: $0.8413$
• Second problem (batteries):
• Z - Score: $z_1=-0.33,z_2 = 0.67$, Percentage/Probability: $0.3779$
• Third problem (gravel):
• (a)(i) Z - Score: $-1.6$, Percentage/Probability: $0.0548$
• (a)(ii) Z - Score: $z_1=-1,z_2 = 1.4$, Percentage/Probability: $0.7605$
• (b) Z - Score: $-0.67$, Weight: $25.47$ kg