HEY CAN ANYONE PLS ANSWER DIS MATH PROBLEM!!!
2 answers:
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Answer:
81.85%
Step-by-step explanation:
Given :
The average summer temperature in Anchorage is 69°F.
The daily temperature is normally distributed with a standard deviation of 7°F .
To Find:What percentage of the time would the temperature be between 55°F and 76°F?
Solution:
Mean =
Standard deviation =
Formula :
Now At x = 55
At x = 76
Now to find P(55<z<76)
P(2<z<-1)=P(z<2)-P(z>-1)
Using z table :
P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185
Now percentage of the time would the temperature be between 55°F and 76°F =
Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.
Answer:
No, we can't reject the dealer's claim with a significance level of 0.05.
Step-by-step explanation:
We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.
You take a random sample of 600 cars in San Francisco and find that 141 are blue.
<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>
SO, Null Hypothesis, : p = 25% {means that 25% of all cars in San Francisco are blue}
Alternate Hypothesis, : p
25% {means that % of all cars in San Francisco who are blue is different from 25%}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of 600 cars in San Francisco who are blue =
= 0.235
n = sample of cars = 600
So, <u><em>test statistics</em></u> =
= -0.866
<em>Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>
Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.