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Suppose that the weight of sweet cherries is normally distributed with mean μ=6 ounces and standard deviation σ=1.4 ounces. What

vazorg [7]

Answer:

The proportion of sweet cherries that weigh less than 5 ounces is 23.88%.

Step-by-step explanation:

We are given that the weight of sweet cherries is normally distributed with mean μ = 6 ounces and standard deviation σ = 1.4 ounces.

Let X = the weight of sweet cherries

SO, X ~ Normal(μ = 6 ounces , σ = 1.4 ounces)

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 6 ounces

$\sigma$ = standard deviation = 1.4 ounces

Now, the proportion of sweet cherries that weigh less than 5 ounces is given by = P(X < 5 ounces)

P(X < 5 ounces) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5-6}{1.4}$ ) = P(Z < -0.71) = 1 - P(Z $\leq$ 0.71)

= 1 - 0.7612 = <u>0.2388</u>

The above probability is calculated by looking at the value of x = 0.71 from the z table which has an area of 0.7612.

8 0

4 years ago

Evaluate the expression -9 W + 9 when w equals 15

makvit [3.9K]

Answer:

The value of the given expression -9w+9 when w=15 is -126

Step-by-step explanation:

Given expression is -9w+9

To find the given expression when w=15:

That is put w=15 in the given expression we get

-9w+9=-9(15)+9

=-135+9 ( adding the like terms )

=-126

Therefore the value of the given expression -9w+9 when w=15 is -126

3 0

3 years ago

Sasha has $55 to spend on clothes. She wants to buy a pair of jeans for $27 and spend the rest on t-shirts. Each t-shirt costs $

Svetradugi [14.3K]

X <span>≤ 10

that's the answer

sorry i need 20 characters lol

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3 0

3 years ago

Read 2 more answers

Help this should be simple

iragen [17]

Answer: I believe it’s D

7 0

3 years ago

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. Whe

kherson [118]

Answer:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: