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2 answers:
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This is a problem of Standard Normal distribution.
We have mean= 12 grams
Standard Deviation = 2.5 grams
First we convert 8.5 to z score. 8.5 converted to z score for given mean and standard deviation will be:
So, from standard normal table we need to find the probability of z score to be less than -1.4. The probability comes out to be 0.0808
Thus, the <span>probability that the strawberry weighs less than 8.5 grams is 0.0808</span>
We have mean= 12 grams
Standard Deviation = 2.5 grams
First we convert 8.5 to z score. 8.5 converted to z score for given mean and standard deviation will be:
So, from standard normal table we need to find the probability of z score to be less than -1.4. The probability comes out to be 0.0808
Thus, the <span>probability that the strawberry weighs less than 8.5 grams is 0.0808</span>
Answer:
The number of business students that must be randomly selected to estimate the mean monthly earnings of business students at one college is 64.
Step-by-step explanation:
The (1 - <em>α</em>) % confidence interval for population mean is:
The margin of error for this interval is:
The information provided is:
<em>σ</em> = $569
MOE = $140
Confidence level = 95%
<em>α</em> = 5%
Compute the critical value of <em>z</em> for <em>α</em> = 5% as follows:
*Use a <em>z</em>-table.
Compute the sample size required as follows:
Thus, the number of business students that must be randomly selected to estimate the mean monthly earnings of business students at one college is 64.