3.. please help!!!!!!!!!!!

Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the popu

tia_tia [17]

Answer:

The minimum head breadth that will fit the clientele is 4.4 inches.

The maximum head breadth that will fit the clientele is 7.8 inches.

Step-by-step explanation:

Let X = head breadths of men that is considered for the helmets.

The random variable X is normally distributed with mean, μ = 6.1 and standard deviation, σ = 1.

To compute the probability of a normal distribution we first need to convert the raw scores to z-scores using the formula:

$z=\frac{x-\mu}{\sigma}$

It is provided that the helmets will be designed to fit all men except those with head breadths that are in the smallest 4.3% or largest 4.3%.

Compute the minimum head breadth that will fit the clientele as follows:

P (X < x) = 0.043

⇒ P (Z < z) = 0.043

The value of z for this probability is:

z = -1.717

*Use a z-table.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\-1.717=\frac{x-6.1}{1}\\x=6.1-(1.717\times 1)\\x=4.383\\x\approx4.4$

Thus, the minimum head breadth that will fit the clientele is 4.4 inches.

Compute the maximum head breadth that will fit the clientele as follows:

P (X > x) = 0.043

⇒ P (Z > z) = 0.043

⇒ P (Z < z) = 1 - 0.043

⇒ P (Z < z) = 0.957

The value of z for this probability is:

z = 1.717

*Use a z-table.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\1.717=\frac{x-6.1}{1}\\x=6.1+(1.717\times 1)\\x=7.817\\x\approx7.8$

Thus, the maximum head breadth that will fit the clientele is 7.8 inches.

5 0

3 years ago

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .